Application of Derivatives: Rate of Change of Quantities

A balloon is at a height of 50 m and is rising at a constant rate of 5 m/s. A cyclist passes beneath it, traveling in a straight line at a constant speed of 10 m/s. How fast is the distance between the cyclist and the balloon increasing 2 seconds later?

Let height of balloon at any instant of time $= h$ meter

Rate of change of height of balloon $= \dfrac{dh}{dt} = 5 \; m/s$

Initial height of balloon $= 50$ meter

$\therefore$ Height of balloon after 2 sec $= h = 50 \; m + 5 \; m/s \times 2 \; s = 60 \; m$

Let distance covered by the cyclist at any instant of time $= x$ meter

Speed of cyclist $= \dfrac{dx}{dt} = 10 \; m/s$

$\therefore$ Distance covered by the cyclist after 2 seconds $= x = 10 \; m/s \times 2 \; s = 20 \; m$

Let the distance between the cyclist and the balloon $= L$ meter

Then, $L = \sqrt{h^2 + x^2}$ $\;\; \cdots$ (1)

Differentiating equation (1) with respect to time t gives

$\dfrac{dL}{dt} = \dfrac{1}{2 \sqrt{h^2 + x^2}} \times \left(2h \dfrac{dh}{dt}+ 2x \dfrac{dx}{dt}\right)$

i.e. $\dfrac{dL}{dt} = \dfrac{1}{\sqrt{h^2 + x^2}} \times \left(h \dfrac{dh}{dt} + x \dfrac{dx}{dt}\right)$ $\;\; \cdots$ (2)

After 2 seconds, $x = 20 \; m$, $h = 60 \; m$, $\dfrac{dx}{dt} = 10 \; m/s$ and $\dfrac{dh}{dt} = 5 \; m/s$

$\therefore$ $\dfrac{dL}{dt} = \dfrac{1}{\sqrt{60^2 + 20^2}} \times \left(60 \times 5 + 20 \times 10\right) = \dfrac{500}{\sqrt{4000}} = \dfrac{500}{20\sqrt{10}} = \dfrac{5\sqrt{10}}{2}$

$\therefore$ After 2 seconds, the distance between the cyclist and the balloon is increasing at the rate of $\dfrac{5 \sqrt{10}}{2} \; m/s$.

Application of Derivatives: Rate of Change of Quantities

The radius of a cylinder is increasing at the rate of 2 cm/s and its altitude is decreasing at the rate of 3 cm/s. Find the rate of change of volume, when the radius is 3 cm and altitude is 5 cm.

Let the radius of cylinder $= r$ cm

Let the height (altitude) of cylinder $= h$ cm

Rate of increase of radius of cylinder $=\dfrac{dr}{dt} = 2 \; cm/s$ $\;\; \cdots$ (1)

Rate of decrease of height of cylinder $= \dfrac{dh}{dt} = -3 \; cm/s$ $\;\; \cdots$ (2)

Volume of cylinder $=V=\pi r^2 h$ $\;\; \cdots$ (3)

Differentiating equation (3) with respect to time t gives

$\dfrac{dV}{dt} = \pi \left(2rh\dfrac{dr}{dt} + r^2 \dfrac{dh}{dt}\right)$ $\;\; \cdots$ (4)

When $r=3 \; cm$ and $h = 5 \; cm$, in view of equations (1) and (2), equation (4) becomes

$\dfrac{dV}{dt} = \pi \left(2 \times 3 \times 5 \times 2 - 3^2 \times 3\right)= 33 \pi$

$\therefore$ The volume of the cylinder is increasing at the rate of $33 \pi \; cm^3/s$.

Application of Derivatives: Rate of Change of Quantities

If $y=7x-x^3$ and x increases at the rate of 4 units per second, how fast is the slope of the curve changing when $x=2$?

Given: Rate at which x increases $= \dfrac{dx}{dt}= 4$ units/s

Slope of the given curve $= \dfrac{dy}{dx}$

Rate of change of slope of the curve $= \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$

The given curve is $y=7x-x^3$

Differentiating the equation of the given curve w.r.t x gives

$\dfrac{dy}{dx} = 7 - 3x^2$

$\therefore$ $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) = \dfrac{d}{dt}\left(7-3x^2\right) = -6x \dfrac{dx}{dt}$

$\therefore$ When $x=2$

$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) = -6 \times 2 \times 4 = -48$

$\therefore$ The slope of the curve is decreasing at the rate 48 units/s.

Application of Derivatives: Rate of Change of Quantities

A ladder 7.5 m long leans against a wall. The ladder slides along the floor away from the wall at the rate of 3 cm/s. How fast is the height of the ladder on the wall decreasing, when the foot of the ladder is 6 m away from the wall?

Length of the ladder $= \ell = 7.5 \; m= 750 \; cm$

Distance of the foot of the ladder from the wall $=x = 6 \; m = 600 \; cm$

Let the height of the ladder on the wall $= h \; cm$

Rate at which the ladder slips away from the wall $= \dfrac{dx}{dt} = 3 \; cm/s$

Now, $\ell = \sqrt{x^2 + h^2}$

$\implies$ $h = \sqrt{\ell^2 - x^2}$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t time t gives

$\begin{aligned} \dfrac{dh}{dt} & = \dfrac{1}{2 \sqrt{\ell^2 - x^2}} \times \left(-2x \dfrac{dx}{dt}\right) \\ & = - \dfrac{600}{\sqrt{750^2 - 600^2}} \times 3 \\ & = -4 \; cm/s \end{aligned}$

$\therefore$ The height of the ladder is decreasing at the rate of 4 cm/s.

Application of Derivatives: Rate of Change of Quantities

A man is moving away from a tower 85 m high at a speed of 4 m/s. Find the rate at which his angle of elevation of the top of the tower is changing when he is at a distance of 60 m from the foot of the tower.

In the figure

OT = tower of height h = 85 m

M = position of the man at any instant of time

x = distance of the man from the tower

$\alpha =$ angle of elevation of the top of the tower as seen by the man

Now, speed at which the man is moving from the tower $= \dfrac{dx}{dt} = 4 \; m/s$

From the figure

$\tan \left(\alpha\right) = \dfrac{h}{x}$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t time t gives

$\sec^2 \left(\alpha\right) \dfrac{d\alpha}{dt} = - \dfrac{h}{x^2} \dfrac{dx}{dt}$

$\begin{aligned} \text{i.e. } \dfrac{d \alpha}{dt} & = -\dfrac{h}{x^2 \sec^2 \left(\alpha\right)} \dfrac{dx}{dt} \\ & = - \dfrac{h}{x^2 \left[1+ \tan^2 \left(\alpha\right)\right]} \dfrac{dx}{dt} \\ & = - \dfrac{h}{x^2 \left[1+ \left(\dfrac{h}{x}\right)^2\right]} \dfrac{dx}{dt} \;\; \left[\text{In view of equation (1)}\right] \\ & = \dfrac{-h}{x^2 + h^2} \dfrac{dx}{dt} \end{aligned}$

Now, when $h = 85 \; m$, $x=60 \; m$ and $\dfrac{dx}{dt} = 4 \; m/s$,

$\dfrac{d\alpha}{dt} = \dfrac{-85}{60^2 + 85^2} \times 4 = -0.0314 \; \text{rad/s}$

$\therefore$ The angle of elevation of the top of the tower is decreasing at the rate of $-0.0314 \; \text{rad/s}$.

Application of Derivatives: Rate of Change of Quantities

A rectangular plate is expanding. Its length x is increasing at the rate 1 cm/s and its width y is decreasing at the rate 0.5 cm/s. At the moment when $x=4$ and $y=3$, find the rate of change of its area, perimeter and diagonal.

Length of rectangular plate $=x \; cm$

Width of rectangular plate $=y \; cm$

Given: $x=4 \; cm; \;\;\; y=3 \; cm$ $\;\; \cdots$ (1)

Rate of increase of length $= \dfrac{dx}{dt} = 1 \; cm/s$ $\;\; \cdots$ (2)

Rate of decrease of width $=\dfrac{dy}{dt} = -0.5 \; cm/s$ $\;\; \cdots$ (3)

Area of the rectangular plate $=A=xy$ $\;\; \cdots$ (4)

Differentiating equation (4) w.r.t time t gives

$\begin{aligned} \dfrac{dA}{dt} & = x \dfrac{dy}{dt} + y \dfrac{dx}{dt} \\ & = 4 \times \left(-0.5\right) + 3 \times 1 \;\; \left[\text{from equations (1), (2) and (3)}\right] \\ & = 1 \; cm^2 / s \end{aligned}$

Perimeter of rectangular plate $=P=2\left(x+y\right)$ $\;\; \cdots$ (5)

Differentiating equation (5) w.r.t time t gives

$\begin{aligned} \dfrac{dP}{dt} & = 2 \left(\dfrac{dx}{dt} + \dfrac{dy}{dt}\right) \\ & = 2 \times \left(1-0.5\right) \;\; \left[\text{from equtions (2) and (3)}\right] \\ & = 1 \; cm/s \end{aligned}$

Diagonal of rectangular plate $= D = \sqrt{x^2 +y^2}$ $\;\; \cdots$ (6)

Differentiating equation (6) w.r.t time t gives

$\begin{aligned} \dfrac{dD}{dt} & = \dfrac{1}{2\sqrt{x^2+y^2}} \times \left(2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}\right) \\ & = \dfrac{1}{\sqrt{4^2 + 3^2}} \times \left[4 \times 1 + 3 \times \left(-0.5\right)\right] \;\; \left[\text{from equations (1), (2) and (3)}\right] \\ & = 0.5 \; cm/s \end{aligned}$

$\therefore$ $\dfrac{dA}{dt}=1 \; cm^2/s; \; \dfrac{dP}{dt}=1 cm/s; \; \dfrac{dD}{dt}=0.5 \; cm/s$

Application of Derivatives: Rate of Change of Quantities

Sand is pouring from a pipe at the rate of $12 \; cm^3 /s$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Let volume of sand pouring out $=V$

Then, rate at which the sand is pouring out $= \dfrac{dV}{dt}=12 \; cm^3/s$ $\;\; \cdots$ (1)

Let radius of the cone formed $=r$

Then height of the cone $=h=\dfrac{r}{6}$

$\therefore$ Radius of cone formed $=r=6h$ $\;\; \cdots$ (2)

Now, volume of cone formed = volume of sand pouring out $=V$

Volume of cone $=V = \dfrac{1}{3}\pi r^2 h$

Substituting the value of r from equation (2) gives

$V = \dfrac{1}{3} \times \pi \times 36h^3$

$\implies$ $V = 12 \pi h^3$ $\;\; \cdots$ (3)

Differentiating equation (3) w.r.t time t gives

$\dfrac{dV}{dt} = 12\pi \times 3h^2 \dfrac{dh}{dt}$

i.e. $\dfrac{dh}{dt} = \dfrac{1}{36\pi h^2}\dfrac{dV}{dt}$ $\;\; \cdots$ (4)

When $h = 4 \; cm$, in view of equation (1), equation (4) becomes

$\dfrac{dh}{dt} = \dfrac{12}{36 \times \pi \times 4^2}$

i.e. $\dfrac{dh}{dt} = \dfrac{1}{48\pi} \; cm/s$

$\therefore$ The height of the sand cone is increasing at the rate of $\dfrac{1}{48\pi} \; cm/s$

Application of Derivatives: Rate of Change of Quantities

A cylinder is heated so that its radius remains twice of its height at any moment. Find the rate of increase of its total surface area and volume, when the radius is 3 cm and the radius increases at the rate 2 cm/s.

Let height of the cylinder $= h \; cm$

Radius of cylinder $= r = 2h \; cm$

$\therefore$ $h = \dfrac{r}{2}$

Given: $r = 3 \; cm$ $\;\; \cdots$ (1)

%$\therefore$ $h = \dfrac{r}{2} = \dfrac{3}{2} \; cm$ $\;\; \cdots$ (2)

Rate of change of radius $= \dfrac{dr}{dt} = 2 \; cm/s$ $\;\; \cdots$ (2)

Total surface area of cylinder $=A=2\pi r \left(r+h\right) = 2 \pi r \left(r+\dfrac{r}{2}\right) = 3 \pi r^2$ $\;\; \cdots$ (3)

Volume of cylinder $=V = \pi r^2 h = \dfrac{1}{2}\pi r^3$ $\;\; \cdots$ (4)

Differentiating equation (3) w.r.t time t gives

$\begin{aligned} \dfrac{dA}{dt} & = 6 \pi r \dfrac{dr}{dt} \\ & = 6 \pi \times 3 \times 2 \;\; \left[\text{from equations (1) and (2)}\right] \\ & = 36 \pi \; cm^2 /s \end{aligned}$

Differentiating equation (4) w.r.t time t gives

$\begin{aligned} \dfrac{dV}{dt} & = \dfrac{3}{2} \pi r^2 \dfrac{dr}{dt} \\ & = \dfrac{3}{2} \pi \times 3^2 \times 2 \;\; \left[\text{from equations (1) and (2)}\right] \\ & = 27 \pi \; cm^3 /s \end{aligned}$

Application of Derivatives: Rate of Change of Quantities

A balloon, which always remains spherical, has a variable diameter $\dfrac{3}{2} \left(2x+1\right)$. Find the rate of change of its volume with respect to x.

Let the diameter of the sphere $=D = \dfrac{3}{2} \left(2x+1\right)$

Volume of sphere $V = \dfrac{4}{3}\pi \times \left(\dfrac{D}{2}\right)^3$

i.e. $V = \dfrac{4\pi}{24} \times D^3$

i.e. $V = \dfrac{\pi}{6} \times \left[\dfrac{3}{2} \left(2x+1\right)\right]^3$

i.e. $V = \dfrac{27\pi}{48}\left(8x^3+12x^2+6x+1\right)$

i.e. $V = \dfrac{9\pi}{16} \left(8x^3+12x^2+6x+1\right)$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t x gives

$\dfrac{dV}{dx}= \dfrac{9\pi}{16}\left(24x^2+24x+6\right)$

i.e. $\dfrac{dV}{dx}= \dfrac{27\pi}{8}\left(4x^2+4x+1\right)$

i.e. $\dfrac{dV}{dx} = \dfrac{27 \pi}{8}\left(2x+1\right)^2$

Application of Derivatives: Rate of Change of Quantities

The surface area of a cube increases at the rate of $12 \; cm^2 /s$. Find the rate at which its volume increases, when its edge has a length of 5 cm.

Let length of edge of cube $=s=5 \; cm$ $\;\; \cdots$ (1)

Let surface area of the cube $=A = 6s^2$

Given: Rate of increase of surface area $=\dfrac{dA}{dt}=12 \; cm^2 /s$ $\;\; \cdots$ (2)

Now $\dfrac{dA}{dt} = 12 s \dfrac{ds}{dt}$ $\implies$ $\dfrac{ds}{dt} = \dfrac{1}{12s} \dfrac{dA}{dt}$ $\; \; \cdots$ (3)

Let volume of cube $=V=s^3 \;\; \cdots$ (4)

Differentiating both sides of equation (4) w.r.t time t gives

$\begin{aligned} \dfrac{dV}{dt} & = 3 s^2 \dfrac{ds}{dt} \\ & = 3s^2 \times \dfrac{1}{12s} \dfrac{dA}{dt} \;\; \left[\text{From equation (3)}\right] \\ & = 3 \times 5^2 \times \dfrac{1}{12 \times 5} \times 12 \; \; \left[\text{From equations (1) and (2)}\right] \\ & = 15 \; cm^3 / s \end{aligned}$

$\therefore$ Rate at which its volume increases $= 15 cm^3 /s$

Application of Derivatives: Rate of Change of Quantities

A particle moves along the curve $6y=x^3+2$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Equation of curve is $6y=x^3+2$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t time t gives

$6 \dfrac{dy}{dt} = 3x^2 \dfrac{dx}{dt}$

i.e. $2 \dfrac{dy}{dt} = x^2 \dfrac{dx}{dt}$ $\;\; \cdots$ (2)

Given: $\dfrac{dy}{dt} = 8 \dfrac{dx}{dt}$ $\;\; \cdots$ (3)

$\therefore$ We have from equations (2) and (3)

$2 \times 8 \dfrac{dx}{dt} = x^2 \dfrac{dx}{dt}$

i.e. $x^2 = 16$

$\implies$ $x=\pm 4$

$\therefore$ Substituting the value of x in equation (1) gives

when $x=+4$, $y=\dfrac{4^3 + 2}{6} = 11$ and

when $x=-4$, $y = \dfrac{\left(-4\right)^3 + 2}{6} = -\dfrac{31}{3}$

$\therefore$ The required points on the curve are $\left(4,11\right)$ and $\left(-4, - \dfrac{31}{3}\right)$

Application of Derivatives: Rate of Change of Quantities

The radius of a cone increases at the rate of 4 cm/sec and the altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of the lateral surface area when the radius is 3 cm and the altitude is 4 cm.

Let $r= 3 \; cm =$ radius of the cone; $h = 4 \; cm =$ height (altitude) of the cone;

Given: $\dfrac{dr}{dt} = 4 \; cm/s$; $\dfrac{dh}{dt} = -3 \; cm/s$

$\ell = \sqrt{r^2 + h^2} =$ slant height of the cone

Lateral surface area of the cone $=A = \pi r \ell = \pi r \sqrt{r^2 + h^2}$

$\begin{aligned} \dfrac{dA}{dt} & = \pi \sqrt{r^2+h^2} \times \dfrac{dr}{dt} + \pi r \times \dfrac{1}{2 \sqrt{r^2 + h^2}} \times \left(2r\dfrac{dr}{dt}+2h\dfrac{dh}{dt}\right) \\ & \\ & = \pi \dfrac{dr}{dt} \left[\sqrt{r^2 + h^2} + \dfrac{r^2}{\sqrt{r^2 + h^2}}\right] + \dfrac{\pi r h}{\sqrt{r^2 + h^2}} \dfrac{dh}{dt} \\ & \\ & = \pi \times 4 \left[\sqrt{3^2 + 4^2}+\dfrac{3^2}{\sqrt{3^2+4^2}}\right] + \dfrac{\pi \times 3 \times 4}{\sqrt{3^2+4^2}} \times \left(-3\right) \\ & \\ & = \dfrac{136 \pi}{5} - \dfrac{36\pi}{5} \\ & \\ & = 20 \pi \; cm^2 / s \end{aligned}$

Application of Derivatives: Approximations

Find the approximate value of $\cos \left(\dfrac{11 \pi}{36}\right)$, assuming the value of $\cos \left(\dfrac{\pi}{3}\right)$.

Let $y = \cos x$ $\;\; \cdots$ (1)

Let $x = \dfrac{\pi}{3}$, $\Delta x = -\dfrac{\pi}{36}$ so that

$\left(x + \Delta x\right) = \dfrac{\pi}{3} - \dfrac{\pi}{36} = \dfrac{11\pi}{36}$ $\;\; \cdots$ (2)

Now $\Delta y = \cos \left(x + \Delta x\right) - \cos x$

i.e. $\Delta y = \cos \left(\dfrac{11\pi}{36}\right) - \cos \left(\dfrac{\pi}{3}\right)$

i.e. $\cos \left(\dfrac{11\pi}{36}\right) = \Delta y + \cos \left(\dfrac{\pi}{3}\right)$

i.e $\cos \left(\dfrac{11 \pi}{36}\right) = \Delta y + \dfrac{1}{2}$ $\;\; \cdots$ (3)

Differentiating equation (1) w.r.t x gives

$\dfrac{dy}{dx} = -\sin x = -\sin \left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}$ $\;\; \cdots$ (4)

Now $\Delta y$ is approximately equal to dy and

$dy = \left(\dfrac{dy}{dx}\right) \times \Delta x = - \dfrac{\sqrt{3}}{2} \times \left(-\dfrac{\pi}{36}\right) = \dfrac{\pi \sqrt{3}}{72}$ [from equations (2) and (4)]

$\therefore$ $\Delta y \approx dy = \dfrac{\pi \sqrt{3}}{72}$ $\;\; \cdots$ (5)

$\therefore$ We have from equations (3) and (5)

$\cos \left(\dfrac{11 \pi}{36}\right) = \dfrac{\pi \sqrt{3}}{72} + \dfrac{1}{2}$

Application of Derivatives: Approximations

The height and the radius of a cylinder are equal. An error of $2\%$ is made in measuring its height. Find the approximate percentage error in its volume.

Let height of the cylinder $= h$

Then radius of the cylinder $= r = h$

Error in measurement of height $= \dfrac{\Delta h}{h} = 2\% = 0.02$

Volume of cylinder $= V = \pi r^2 h = \pi h^3$

$\therefore$ $\dfrac{dV}{dh} = 3 \pi h^2$

Now, $\Delta V = \left(\dfrac{dV}{dh}\right) \times \Delta h$

$\begin{aligned} \therefore \% \text{Error in volume} & = \dfrac{\Delta V}{V} \times 100 \% \\ & = \dfrac{\left(\dfrac{dV}{dh}\right) \times \Delta h}{V} \times 100 \% \\ & = \dfrac{3 \pi h^2 \times \Delta h}{\pi h^3} \times 100 \% \\ & = 3 \times \dfrac{\Delta h}{h} \times 100 \% \\ & = 3 \times 0.02 \times 100 \% = 6 \% \end{aligned}$

Application of Derivatives: Approximations

Find the percentage error in calculating the volume of a cubical box, if an error of $1\%$ is made in measuring the length of edges of the cube.

Let the length of edge of cube $= \ell$

Given: $\dfrac{\Delta \ell}{\ell} = 1 \% = 0.01$

Now, volume of cube $= V = \ell^3$

$\therefore$ $\dfrac{dV}{d\ell} = 3\ell^2$

Now, $\Delta V = \left(\dfrac{dV}{d\ell}\right) \times \Delta \ell$

$\begin{aligned} \% \text{Error in V} = \dfrac{\Delta V}{V} \times 100 \% & = \dfrac{\left(dV/d \ell\right) \times \Delta \ell}{V} \times 100 \% \\ & = \dfrac{3 \ell^2 \times \Delta \ell}{\ell^3} \times 100 \% \\ & = 3 \times \left(\dfrac{\Delta \ell}{\ell}\right) \times 100 \% \\ & = 3 \times 0.01 \times 100 \% \\ & = 3 \% \end{aligned}$

Application of Derivatives: Approximations

The time period of a simple pendulum is given as $T = 2\pi \sqrt{\dfrac{\ell}{g}}$. If there is a $4\%$ error in measuring the period of the pendulum, find the approximate percentage error in the measurement of length $\ell$ of the pendulum. g is the acceleration due to gravity, a constant.

$\dfrac{\Delta T}{T} = 4\% = \dfrac{4}{100}$

$T = 2\pi \sqrt{\dfrac{\ell}{g}}$

$\therefore$ $\log T = \log \left(2\pi\right) + \dfrac{1}{2} \log \left(\ell\right) - \dfrac{1}{2} \log \left(g\right)$

$\therefore$ $\dfrac{1}{T} \times \dfrac{dT}{d\ell} = \dfrac{1}{2} \times \dfrac{1}{\ell}$

i.e. $\dfrac{dT}{d\ell} = \dfrac{T}{2 \ell} \;\; \cdots$ (1)

Now, $\Delta T = \left(\dfrac{dT}{d\ell}\right) \times \Delta \ell$

$\implies$ $\Delta \ell = \dfrac{\Delta T}{\left(dT / d\ell\right)}$

$\begin{aligned} \therefore \% \text{Error in } \ell = \dfrac{\Delta \ell}{\ell} \times 100 \% & = \dfrac{\Delta T}{\ell \times \left(dT / d\ell\right)} \times 100 \% \\ & = \dfrac{\Delta T \times 2 \ell}{\ell \times T} \times 100 \% \;\; \left[\text{From equation (1)}\right] \\ & = 2 \times \dfrac{\Delta T}{T} \times 100 \% \\ & = 2 \times \dfrac{4}{100} \times 100 = 8 \% \end{aligned}$

Application of Derivatives: Approximations

If in a triangle ABC, the side a and the angle A remain constant, while the remaining elements are changed slightly, show that $\dfrac{db}{\cos B} + \dfrac{dc}{\cos C} = 0$

According to sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k \left(\text{ constant}\right)$

$\therefore$ $a = k \sin A$, $b = k \sin B$, $c = k \sin C$

$\therefore$ $\dfrac{da}{dA}= k \cos A$, $\dfrac{db}{dB} = k \cos B$, $\dfrac{dc}{dC} = k \cos C$

But $\dfrac{da}{dA} = 0$ since a is a constant (given)

Now $db = \left(\dfrac{db}{dB}\right) \times \Delta B = k \cos B \times \Delta B$

$\implies$ $\dfrac{db}{\cos B} = k \Delta B$

and $dc = \left(\dfrac{dc}{dC}\right) \times \Delta C = k \cos C \times \Delta C$

$\implies$ $\dfrac{dc}{\cos C} = k \Delta C$

$\begin{aligned} \therefore \dfrac{db}{\cos B} + \dfrac{dc}{\cos C} & = k \Delta B + k \Delta C \\ & = k \left(\Delta B + \Delta C\right) \\ & = k \Delta \left(B+C\right) \\ & = k \Delta \left(\pi - A\right) \;\; \left[\text{Note: In a triangle, }A+B+C=\pi\right] \\ & = k \times 0 \;\; \left[\text{Note: A is a constant (given)}\right] \\ & = 0 \end{aligned}$

$\therefore$ $\dfrac{db}{\cos B}+ \dfrac{dc}{\cos C} = 0$

Application of Derivatives: Approximations

Area of a triangle is given by $A = \dfrac{1}{2}ab \sin C$ where the symbols have their usual meanings. If $C = \dfrac{\pi}{3}$ and there is an error in measuring C by $x \%$, calculate the approximate percentage error in the area if a and b are constant.

Given: $C=\dfrac{\pi}{3}$; $\dfrac{\Delta C}{C} = x\% $ $\implies$ $\Delta C = \dfrac{x}{100} \times C = \dfrac{\pi x}{300}$

Area $A = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}ab \sin \left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{4}ab$

$\dfrac{dA}{dC} = \dfrac{1}{2}ab \cos C$

$\begin{aligned} \text{Now, } \Delta A & = \left(\dfrac{dA}{dC}\right)\times \Delta C \\ & = \dfrac{1}{2}ab \cos C \times \dfrac{\pi x}{300} \\ & = \dfrac{1}{2}ab \times \cos \left(\dfrac{\pi}{3}\right) \times \dfrac{\pi x}{300} \\ & = \dfrac{1}{4} ab \times \dfrac{\pi x}{300} \end{aligned}$

$\begin{aligned} \therefore \% \; \text{Error in area} & = \dfrac{\Delta A}{A} \times 100 \; \% \\ & = \dfrac{\dfrac{1}{4}ab \times \dfrac{\pi x}{300}}{\dfrac{\sqrt{3}}{4}ab} \times 100 \; \% \\ & = \dfrac{\pi x}{3\sqrt{3}} \; \% \end{aligned}$

Application of Derivatives: Approximations

Radius of a circle is 15 cm and it increases by $3\%$. Find the approximate increase in its area using differentials.

Radius of circle $=r=15$ cm

$\Delta r = 3 \% = 0.03$

Area of circle $=A=\pi r^2$

$\therefore$ $\dfrac{dA}{dr}= 2 \pi r$

$\begin{aligned} \text{Increase in area} = \Delta A & = \left(\dfrac{dA}{dr}\right) \times \Delta r \\ & = 2 \pi r \times \Delta r \\ & = 2 \times \pi \times 15 \times 0.03 \\ & = 0.9 \pi \;\; cm^2 \end{aligned}$

Application of Derivatives: Approximations

Kinetic energy is given by $k = \dfrac{1}{2}mv^2$. For a constant mass, there is approximately 1% increase in energy. Find the increase in velocity v which causes this increase in energy.

$k = \dfrac{1}{2}mv^2$

$\therefore$ $\dfrac{dk}{dv}=mv$

$\Delta k = 1\% \text{ of k}=0.01 \times \dfrac{1}{2}mv^2 = 0.005 mv^2$

Now, $\Delta k = \left(\dfrac{dk}{dv}\right) \times \Delta v$

$\therefore$ $\Delta v = \dfrac{\Delta k}{dk / dv}= \dfrac{0.005 mv^2}{mv} = 0.005 v$

i.e. velocity increases by $0.5 \%$

Application of Derivatives: Approximations

The pressure p and the volume V of a gas are related as $pV^{1/4} = a$, constant. Find the percentage increase in the volume corresponding to a diminution of $\dfrac{1}{2} \%$ in the pressure.

Percentage decrease in pressure $= \dfrac{1}{2} \%$

i.e. $\dfrac{\Delta p}{p} = -\dfrac{1}{2} \% = -\dfrac{1}{200}$ $\;\; \cdots$ (1)

Now, $pV^{1/4} = a$

$\therefore$ $\log p + \dfrac{1}{4} \log V = \log a = 0$ [since a is a constant, $\log a =0$]

i.e. $\log V = -4 \log p$ $\;\; \cdots$ (2)

Differentiating equation (2) w.r.t p gives

$\dfrac{1}{V} \dfrac{dV}{dp} = -\dfrac{4}{p}$

i.e. $\dfrac{dV}{dp} = -\dfrac{4V}{p}$ $\;\; \cdots$ (3)

Now, $\Delta V = \left(\dfrac{dV}{dp}\right) \times \Delta p = -\dfrac{4V}{p} \times \Delta p = -4V \times \left(\dfrac{\Delta p}{p}\right) = -4V \times \left(-\dfrac{1}{200}\right) = \dfrac{V}{50}$

Now, $\%$ Increase in volume $= \dfrac{\Delta V}{V} \times 100 \% = \dfrac{1}{V} \times \dfrac{V}{50} \times 100 \% = 2 \%$

Application of Derivatives: Approximations

If the radius of a cone is twice its height, find the approximate error in the calculation of its volume, when the radius is 10 cm and the error in radius is 0.01 cm

Let the height of cone $=h$ cm

Then radius of cone $=r = 2h$ cm $\implies$ $h = \dfrac{r}{2}$ cm

Let the error in radius $=\Delta r = 0.01$ cm

Volume of cone $= V = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \pi r^2 \times \dfrac{r}{2} = \dfrac{1}{6}\pi r^3$

$\therefore$ $\dfrac{dV}{dr} = \dfrac{\pi r^2}{2}$

$\begin{aligned} \therefore \text{Error in volume } \Delta V & = \left(\dfrac{dV}{dr}\right)\times \Delta r \\ & = \left(\dfrac{\pi r^2}{2}\right) \times \Delta r \\ & = \dfrac{\pi \times 10^2}{2} \times 0.01 \\ & = \dfrac{\pi}{2} =1.571 \;\; \text{cm}^3 \end{aligned}$

Application of Derivatives: Approximations

Find the approximate value of $f\left(2.01\right)$, where $f\left(x\right) = 4x^2+5x+2$

Let $x=2$, $\Delta x = 0.01$ so that $x + \Delta x = 2.01$

$f\left(x\right) = f\left(2\right) = 4 \times 2^2 + 5 \times 2 + 2 = 28$

$f'\left(x\right) = 8x + 5$

$\begin{aligned} \text{Now}, f\left(x+\Delta x\right) & = f\left(x\right) + \Delta f \left(x\right) \\ & = f \left(x\right) + f'\left(x\right) \cdot \Delta x \\ & = \left(4x^2 + 5x + 2\right) + \left(8x + 5\right) \times \Delta x \end{aligned}$

$\begin{aligned} \therefore \; f\left(2.01\right) & = 28 + \left(8 \times 2 + 5\right) \times 0.01 \\ & = 28 + 0.21 \\ & = 28.21 \end{aligned}$

Application of Derivatives: Approximations

Using differentials, find the approximate value of $\sqrt{51}$

$\sqrt{51}=\left(51\right)^{\frac{1}{2}}$

Let $y = x^{\frac{1}{2}}$

Let $x = 49$, $\Delta x = 2$ so that $x + \Delta x = 51$

Now, $\Delta y = \left(x + \Delta x\right)^{\frac{1}{2}} - x^{\frac{1}{2}}$

i.e. $\Delta y = \left(51\right)^{\frac{1}{2}} - \left(49\right)^{\frac{1}{2}}= \left(51\right)^{\frac{1}{2}}-7$

$\therefore$ $\left(51\right)^{\frac{1}{2}} = \Delta y + 7$ $\;\; \cdots$ (1)

Also $\dfrac{dy}{dx}= \dfrac{1}{2 \sqrt{x}}$

Now, $\Delta y$ is approximately equal to dy and

$dy = \left(\dfrac{dy}{dx}\right) \Delta x = \dfrac{1}{2 \sqrt{x}} \times \Delta x = \dfrac{1}{2 \sqrt{49}} \times 2 = \dfrac{1}{7}= 0.14286$

$\therefore$ Approximate value of $\Delta y = dy = 0.14286$

$\therefore$ We have from equation (1)

$\sqrt{51} = 0.14286 + 7 = 7.14286$

Application of Derivatives: Tangents and Normals

For the curve $y=x^2+3x+4$, find all the points at which the tangent passes through the origin.

$y=x^2+3x+4$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t x gives

$\dfrac{dy}{dx}=2x+3$

The required tangent passes through the origin $\left(0,0\right)$

$\therefore$ $\;$ Equation of tangent is

$y-0 = \left(2x+3\right) \left(x-0\right)$

i.e. $y=2x^2 + 3x$ $\;\; \cdots$ (2)

Solving equations (1) and (2) simultaneously gives

$2x^2 + 3x = x^2 + 3x + 4$

i.e. $x^2 = 4$ $\implies$ $x = \pm 2$

Substituting the value of x in equation (1) gives

when $x=+2$, $y= \left(2\right)^2 + 3 \times 2 + 4 = 14$

when $x=-2$, $y = \left(-2\right)^2 + 3 \times \left(-2\right) + 4 = 2$

$\therefore$ $\;$ The required points are $\left(2,14\right)$ and $\left(-2,2\right)$

Application of Derivatives: Tangents and Normals

Find the measure of angle between the curves $x^2 - y^2 =3$ and $x^2 + y^2 -4x +3 = 0$

$x^2 -y^2 =3$ $\;\; \cdots$ (1)

$x^2 + y^2 -4x + 3 = 0$ $\;\; \cdots$ (2)

Adding equations (1) and (2) gives

$2x^2 -4x=0$

i.e. $x \left(x-2\right)=0$ $\implies$ $x=0$ or $x =2$

$\therefore$ $\;$ From equation (1),

when $x=0$, $y^2 = -3$

$\therefore$ $\;$ $x=0$ is not a valid answer.

when $x=2$, $y^2 = 1$ $\implies$ $y = \pm 1$

$\therefore$ $\;$ The two curves intersect at the points $\left(2,1\right)$ and $\left(2,-1\right)$

Differentiating equation (1) w.r.t x gives

$2x -2 y \dfrac{dy}{dx}=0$

i.e. $\dfrac{dy}{dx}=\dfrac{x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 2 = m_1 \left(\text{say}\right)$

and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=-2 = m_3 \left(\text{say}\right)$

Differentiating equation (2) w.r.t x gives

$2x + 2y \dfrac{dy}{dx}-4=0$

i.e. $\dfrac{dy}{dx}= \dfrac{2-x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 0 = m_2 \left(\text{say}\right)$

and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=0 = m_4 \left(\text{say}\right)$

Let $\alpha$ be the angle between the two tangents at the point $\left(2,1\right)$

Then $\tan \alpha = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right|= \left|\dfrac{2-0}{1+2\times0}\right|=2$

$\implies$ $\alpha = \tan^{-1}\left(2\right)$

Let $\beta$ be the angle between the two tangents at the point $\left(2,-1\right)$

Then $\tan \beta = \left|\dfrac{m_3 - m_4}{1+m_3 m_4}\right|= \left|\dfrac{-2-0}{1+\left(-2\right)\times0}\right|=2$

$\implies$ $\beta = \tan^{-1}\left(2\right)$

Now, angle between the tangents of curves = angle between the curves

$\therefore$ Angle between the two curves $=\tan^{-1}\left(2\right)$

Application of Derivatives: Tangents and Normals

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta + a \theta \sin \theta$, $y = a \sin \theta - a \theta \cos \theta$ is at a constant distance from the origin.

$x = a \cos \theta + a \theta \sin \theta$

$\therefore$ $\;$ $\dfrac{dx}{d\theta}=-a\sin \theta + a \sin \theta + a \theta \cos \theta$

i.e. $\dfrac{dx}{d\theta} = a \theta \cos \theta$ $\;\; \cdots$ (1)

$y = a \sin \theta - a \theta \cos \theta$

$\therefore$ $\;$ $\dfrac{dy}{d\theta} = a \cos \theta -a \cos \theta + a \theta \sin \theta$

i.e. $\dfrac{dy}{d\theta} = a \theta \sin \theta$ $\;\; \cdots$ (2)

$\therefore$ $\;$ $\dfrac{dy}{dx}= \dfrac{dy /d\theta}{dx / d \theta} = \dfrac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$ $\;\;$ [from equations (1) and (2)]

$\therefore$ $\;$ Slope of normal $= \dfrac{-1}{dy / dx}=-\cot \theta$

$\therefore$ $\;$ Equation of normal is

$y - \left(a \sin \theta - a \theta \cos \theta\right) = -\cot \theta \left(x-a\cos \theta -a \theta \sin \theta\right)$

i.e. $\sin \theta\left(y-a\sin \theta + a\theta cos \theta\right) = -\cos \theta \left(x-a\cos\theta - a \theta \sin \theta\right)$

i.e. $y \sin \theta -a \sin^2 \theta + a \theta \sin \theta \cos \theta = -x \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$

i.e. $x \cos \theta + y \sin \theta=a \left(\sin^2 \theta + \cos^2 \theta\right)$

i.e. $x \cos \theta + y \sin \theta - a = 0$

$\therefore$ $\;$ Distance of the normal from the origin $= \left|\dfrac{-a}{\sqrt{\cos^2 \theta + \sin^2 \theta}}\right|=a$

$\therefore$ $\;$ The normal is at a constant distance from the origin.

Note: Distance of the line $Ax+By+C=0$ from the point $\left(x_0,y_0\right)$ is $\left|\dfrac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}\right|$

Application of Derivatives: Tangents and Normals

Show that the curves $xy=a^2$ and $x^2+y^2=2a^2$ touch each other.

The two curves are

$xy=a^2$ $\implies$ $x = \dfrac{a^2}{y}$ $\;\; \cdots$ (1)

$x^2 + y^2 = 2a^2$ $\implies$ $\left(\dfrac{a^2}{y}\right)^2 + y^2 = 2a^2$ [in view of equation (1)]

i.e. $a^4 + y^4 = 2a^2 y^2$

i.e. $y^4 - 2a^2 y^2 +a^4 =0$

i.e. $\left(y^2-a^2\right)^2 = 0$

i.e. $y^2 -a^2 =0$ $\implies$ $y^2 = a^2$ $\implies$ $y = \pm a$

$\therefore$ $\;$ From equation (1),

when $y = a$, $x=a$ and

when $y = -a$, $x=-a$

$\therefore$ $\;$ The given curves intersect each other at the points $\left(a,a\right)$ and $\left(-a,-a\right)$

Differentiating $xy=a^2$ w.r.t x gives

$x \dfrac{dy}{dx}+y=0$

i.e. $\dfrac{dy}{dx}=\dfrac{-y}{x}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1=m_1 \left(\text{say}\right)$ $\;\; \cdots$ (2)

Differentiating $x^2 + y^2 = 2a^2$ w.r.t x gives

$2x+2y\dfrac{dy}{dx}=0$

i.e. $\dfrac{dy}{dx}=\dfrac{-x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1 = m_2 \left(\text{say}\right)$ $\;\; \cdots$ (3)

Let $\phi$ be the angle between the two tangents.

Then $\tan \phi = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right| = \left|\dfrac{-1+1}{1+\left(-1\right)\left(-1\right)}\right|=0$

$\implies$ $\phi = 0$

i.e. the two tangents touch each other.

$\implies$ The two curves touch each other.

Application of Derivatives: Tangents and Normals

Show that the curves $4x=y^2$ and $4xy=k$ cut at right angle if $k^2=512$

Since the curves cut at right angle $\implies$ their tangents are perpendicular to each other

Consider $4x=y^2$

Differentiating w.r.t x gives

$4= 2y \dfrac{dy}{dx}$

$\implies$ $\dfrac{dy}{dx}=\dfrac{2}{y}$

Consider $4xy=k$

Differentiating w.r.t x gives

$4\left(x\dfrac{dy}{dx}+y\right)=0$

$\implies$ $\dfrac{dy}{dx}=-\dfrac{y}{x}$

Since the tangents are perpendicular to each other

$\implies$ $\dfrac{2}{y}\times \left(\dfrac{-y}{x}\right)=-1$ $\implies$ $x=2$

Substituting the value of x in the equation of the curve $4xy=k$ gives

$8y=k$ $\implies$ $y = \dfrac{k}{8}$

Substituting the values of x and y in the equation of the curve $4x=y^2$ gives

$4 \times 2 = \left(\dfrac{k}{8}\right)^2$

$\implies$ $k^2=8\times 64 = 512$

$\therefore$ $\;$ The two curves cut at right angle if $k^2 = 512$

Application of Derivatives: Tangents and Normals

Find the equation of tangent to the curve $x^2-2y^2=8$ which are perpendicular to the line $x-y+29=0$

Slope of the line $x-y+29=0$ is $m=1$

Since the required tangent is perpendicular to the given line

slope of tangent $=-\dfrac{1}{m}=-1$ $\;\; \cdots$ (1)

Differentiating the equation of the curve $x^2-2y^2=8$ with respect to x gives

$2x-4y \dfrac{dy}{dx}=0$

i.e. $2y \dfrac{dy}{dx}=x$

$\implies$ $\dfrac{dy}{dx}=\dfrac{x}{2y}$ $\;\; \cdots$ (2)

$\therefore$ $\;$ We have from equations (1) and (2)

$\dfrac{x}{2y}=-1$

$\implies$ $x=-2y$ $\;\; \cdots$ (3)

In view of equation (3), the equation of the curve becomes

$\left(-2y\right)^2-2y^2=8$

i.e. $2y^2=8$ $\implies$ $y^2=4$ $\implies$ $y=\pm 2$

Substituting the value of y in equation (3) gives

$x=-2y=\mp 4$

$\therefore$ $\;$ The tangents to the given curve are located at the points $\left(-4,2\right)$, $\left(4,-2\right)$

$\therefore$ $\;$ Equation of tangent at $\left(-4,2\right)$ is

$y-2=-1\left(x+4\right)$ $\implies$ $x+y+2=0$

$\therefore$ $\;$ Equation of tangent at $\left(4,-2\right)$ is

$y+2=-1\left(x-4\right)$ $\implies$ $x+y-2=0$

$\therefore$ $\;$ The two tangents to the given curve are $x+y+2=0$ and $x+y-2=0$

Application of Derivatives: Tangents and Normals

Find the equation of tangent to $x=\cos \theta$, $y=\sin \theta$ $\;\;$ $\theta \in \left[0,2\pi\right)$ at $\theta=\dfrac{\pi}{4}$

$\dfrac{dx}{d\theta}=-\sin \theta$

$\dfrac{dy}{d\theta}=\cos \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}=\dfrac{dy / d\theta}{dx / d\theta} = -\dfrac{\cos \theta}{\sin \theta}=-\cot \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{4}} = -\cot \left(\dfrac{\pi}{4}\right)=-1$

When $\theta = \dfrac{\pi}{4}$,

$x = \cos \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$y = \sin \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$\therefore$ $\;\;$ Equation of tangent at $\theta=\dfrac{\pi}{4}$ is

$y-\dfrac{1}{\sqrt{2}}=-1 \left(x-\dfrac{1}{\sqrt{2}}\right)$

i.e. $x\sqrt{2} + y \sqrt{2}=2$

Application of Derivatives: Tangents and Normals

The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$ and cuts the Y axis at a point where its gradient is 3. Find a, b and c.

When the curve $y=ax^3+bx^2+cx+5$ cuts the Y axis, the X coordinate of the point is 0.

$\implies$ $y=5$

$\therefore$ $\;\;$ At the point $\left(0,5\right)$, the gradient of the curve is 3

Differentiating the curve $y=ax^3+bx^2+cx+5$ with respect to x gives

$\dfrac{dy}{dx}=3ax^2+2bx+c \;\; \cdots$ (1)

$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\left(0,5\right)}=c$

The gradient of the curve at $\left(0,5\right)$ is 3

$\implies$ $c=3 \;\; \cdots$ (2)

The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$

$\implies$ $0=\left(-2\right)^3 a + \left(-2\right)^2 b -2c+5$

i.e. $8a-4b+2c = 5$

Substituting the value of c gives

$8a-4b=-1 \;\; \cdots$ (3)

Since the curve touches the X axis at $\left(-2,0\right)$

$\implies$ X axis is tangent to the curve at $\left(-2,0\right)$

$\implies$ $\dfrac{dy}{dx}\bigg|_{\left(-2,0\right)}=0$

$\therefore$ $\;\;$ We have from equations (1) and (2)

$3 \times \left(-2\right)^2 a + 2 \times \left(-2\right)b + 3 = 0$

i.e. $12a-4b=-3 \;\; \cdots$ (4)

Solving equations (3) and (4) simultaneously gives

$4a = -2$ $\implies$ $a = -\dfrac{1}{2}$

Substituting the value of a in equation (3) gives

$8 \times \left(\dfrac{-1}{2}\right)-4b=-1$

i.e. $4b = -3$ $\implies$ $b = -\dfrac{3}{4}$

$\therefore$ $\;\;$ $a=-\dfrac{1}{2}$, $b = -\dfrac{3}{4}$, $c=3$

Application of Derivatives: Tangents and Normals

Find the equation of the tangent to the curve $y=\sqrt{3x-2}$ which is parallel to the line $4x-2y+5=0$

Differentiating the given curve w.r.t x gives

$\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-2}}$ $\;\; \cdots$ (1)

Slope of the line $4x-2y+5=0$ is 2 $\cdots$ (2)

Since the tangent to the curve is parallel to the given line

$\implies$ Slope of tangent $=$ Slope of the line

$\therefore$ $\;$ From equations (1) and (2) we have

$\dfrac{3}{2\sqrt{3x-2}} = 2$

i.e. $3x-2 = \dfrac{9}{16}$ $\implies$ $x = \dfrac{41}{48}$

Substituting the value of x in the equation of curve gives

$y = \sqrt{\dfrac{123}{48}-2} = \sqrt{\dfrac{123-96}{48}}=\sqrt{\dfrac{27}{48}} = \dfrac{3}{4}$

$\therefore$ $\;$ From equation (1), slope of tangent at $\left(\dfrac{41}{48}, \dfrac{3}{4}\right)$ is

$\dfrac{dy}{dx} \bigg|_{\left(\frac{41}{48}, \frac{3}{4}\right)} = \dfrac{3}{2\sqrt{\left(3\times \dfrac{41}{48}\right)-2}} = \dfrac{3}{2 \sqrt{\dfrac{9}{16}}}=2$

$\therefore$ $\;$ Equation of tangent to the given curve parallel to the line $4x-2y+5=0$ is

$y - \dfrac{3}{4} = 2 \left(x - \dfrac{41}{48}\right)$

i.e. $4y-3=8 \left(\dfrac{48x-41}{48}\right)$

i.e. $24y - 18 = 48x - 41$

i.e. $48x - 24y -23=0$

Application of Derivatives: Tangents and Normals

Find the points on the curve $x^2+y^2-2x-3=0$ at which the tangents are parallel to the X axis.

Differentiating the given curve w.r.t x gives

$2x+2y\dfrac{dy}{dx}-2=0$

i.e. $\dfrac{dy}{dx}=1-x$

Since the tangents are parallel to the X axis $\implies$ Slope of tangent $=0$

i.e. $\dfrac{dy}{dx}=0$

i.e. $1-x=0$ $\implies$ $x=1$

Substituting the value of x in the equation of curve gives

$1+y^2 -2 - 3 =0$

i.e. $y^2 = 4$ $\implies$ $y = \pm 2$

$\therefore$ $\;$ The required points are $\left(1,2\right)$ and $\left(1,-2\right)$

Application of Derivatives: Tangents and Normals

Find the equations of the tangent and normal to the curve $y^2 = \dfrac{x^3}{4-x}$ at the point $\left(2,-2\right)$

$y^2 = \dfrac{x^3}{4-x}$

Differentiating w.r.t x gives

$2y \dfrac{dy}{dx} = \dfrac{\left(4-x\right)\times 3x^2 -x^3 \times \left(-1\right)}{\left(4-x\right)^2}$

i.e. $\dfrac{dy}{dx} = \dfrac{1}{2y} \left[\dfrac{12x^2-2x^3}{\left(4-x\right)^2}\right]$

i.e. $\dfrac{dy}{dx} = \dfrac{1}{y} \left[\dfrac{6x^2 -x^3}{\left(4-x\right)^2}\right]$

$\therefore$ $\;$ $\dfrac{dy}{dx} \bigg|_{\left(2,-2\right)} = \dfrac{-1}{2}\left[\dfrac{\left(6\times 2^2\right)-2^3}{\left(4-2\right)^2}\right]$

i.e. $\dfrac{dy}{dx} \bigg|_{\left(2,-2\right)} = \dfrac{-1}{2} \times \dfrac{16}{4}=-2$

$\therefore$ $\;$ Slope of tangent at $\left(2,-2\right) = -2$

$\therefore$ $\;$ Equation of tangent at $\left(2,-2\right)$ is $\;\;$ $y+2 = -2 \left(x-2\right)$

i.e. $2x+y-2=0$

Slope of normal at $\left(2,-2\right)$ $= \dfrac{-1}{\text{slope of tangent}} = \dfrac{1}{2}$

$\therefore$ $\;$ Equation of normal at $\left(2,-2\right)$ is $\;\;$ $y+2 = \dfrac{1}{2} \left(x-2\right)$

i.e. $x-2y-6=0$

Application of Derivatives: Tangents and Normals

Find the slope of the normal to the curve $x=1-a \sin \theta$, $y=b\cos^2 \theta$ at $\theta = \dfrac{\pi}{2}$

$x = 1-a\sin \theta$

$\therefore$ $\;$ $\dfrac{dx}{d\theta} = -a \cos \theta$

$y = b \cos^2 \theta$

$\therefore$ $\;$ $\dfrac{dy}{d\theta}=-2b\cos \theta \sin \theta$

$\therefore$ $\;$ Slope of tangent to the given curve $= \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$

i.e. $\dfrac{dy}{dx} = \dfrac{-2b\cos \theta \sin \theta}{-a\cos \theta}$

i.e. $\dfrac{dy}{dx} = \dfrac{2b}{a} \sin \theta$

$\therefore$ $\;$ $\dfrac{dy}{dx} \bigg|_{\theta = \frac{\pi}{2}} = \dfrac{2b}{a} \sin \left(\dfrac{\pi}{2}\right) = \dfrac{2b}{a}$

Slope of normal to the curve $\left(\text{at }\theta = \dfrac{\pi}{2}\right)$ $= \dfrac{-1}{\text{slope of tangent at }\theta = \frac{\pi}{2}}$

$\therefore$ $\;$ Slope of normal $=-\dfrac{a}{2b}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 1} \; \left(\dfrac{1}{\log x}-\dfrac{1}{x-1}\right)$

$\begin{aligned} & \lim\limits_{x \to 1} \; \left(\dfrac{1}{\log x}-\dfrac{1}{x-1}\right) & \left(\infty - \infty \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x-1-\log x}{\left(x-1\right)\log x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{\dfrac{d}{dx}\left[x-1-\log x\right]}{\dfrac{d}{dx}\left[\left(x-1\right)\log x\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{1-0-\dfrac{1}{x}}{\dfrac{x-1}{x}+\log x} & \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x-1}{x-1+\log x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{\dfrac{d}{dx}\left(x-1\right)}{\dfrac{d}{dx}\left(x-1+\log x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{1}{1+\dfrac{1}{x}} & \\ & & \\ & = \lim\limits_{x \to 1} \; \dfrac{x}{x+1} = \dfrac{1}{1+1} = \dfrac{1}{2} & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\left(\tan^{-1}x\right)^2}{\log \left(1+x^2\right)}$

$\begin{aligned} &\lim\limits_{x \to 0} \; \dfrac{\left(\tan^{-1}x\right)^2}{\log \left(1+x^2\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\tan^{-1}x\right)^2}{\dfrac{d}{dx}\left[\log \left(1+x^2\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2 \tan^{-1}x / \left(1+x^2\right)}{2x / \left(1+x^2\right)} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\tan^{-1}x}{x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\tan^{-1}x\right)}{\dfrac{d}{dx}\left(x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{1/1+x^2}{1} & \\ & & \\ & = \dfrac{1}{1+0} = 1 & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{e^x \sin x - x -x^2}{x^2 + x \log \left(1-x\right)}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{e^x \sin x -x -x^2}{x^2 + x \log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \sin x -x - x^2\right\}}{\dfrac{d}{dx}\left\{x^2 + x \log \left(1-x\right)\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x + e^x \sin x -1 - 2x}{2x-\dfrac{x}{1-x}+\log \left(1-x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{e^x \cos x + e^x \sin x -1 -2x\right\}}{\dfrac{d}{dx}\left\{2x - \dfrac{x}{1-x}+ \log \left(1-x\right) \right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x \cos x -e^x \sin x + e^x \sin x + e^x \cos x -2}{2 - \dfrac{1-x+x}{\left(1-x\right)^2}- \dfrac{1}{1-x}} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2e^x \cos x -2}{2 - \dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left\{2e^x \cos x -2\right\}}{\dfrac{d}{dx}\left\{2-\dfrac{1}{\left(1-x\right)^2}-\dfrac{1}{1-x}\right\}} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2\left(e^x \cos x -e^x \sin x\right)}{\dfrac{-2}{\left(1-x\right)^3}- \dfrac{1}{\left(1-x\right)^2} } & \\ & & \\ & = \dfrac{2\times \left(1-0\right)}{-2-1} & \\ & & \\ & = - \dfrac{2}{3} \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x }$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(x^2 e^x\right)}{\dfrac{d}{dx}\tan^2 x} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{x^2 e^x + 2x e^x}{2 \tan x \sec^2 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x \left(x+2\right)}{2 \tan x \sec^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left[xe^x \left(x+2\right)\right]}{\dfrac{d}{dx}\left(2 \tan x \sec^2 x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x + \left(x+2\right)\left(xe^x + e^x\right)}{2 \left(\tan x \times 2\sec x \times \sec x \tan x + \sec^2 x \times \sec^2 x\right) } & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4xe^x + x^2 e^x + 2e^x}{4 \tan^2 x \sec^2 x + 2 \sec^4 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x\left(x^2+4x+2\right)}{4\sec^2 x \left(\sec^2 x -1\right)+2\sec^4 x} & \\ & & \\ & = \dfrac{1\left(0+0+2\right)}{4 \times 1 \times \left(1-1\right)+ 2 \times 1} = \dfrac{2}{2} = 1 \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right)$

$\begin{aligned} & \lim\limits_{x \to 1^{-}} \; \log \left(1-x\right) \cot \left(\dfrac{\pi x}{2}\right) & \left(\infty-0 \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\log \left(1-x\right)}{\tan \left(\dfrac{\pi x}{2}\right)} & \left(\dfrac{\infty}{\infty} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\log \left(1-x\right)\right]}{\dfrac{d}{dx}\left[\tan \left(\dfrac{\pi x}{2}\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{-1}{1-x}}{\dfrac{\pi}{2}\sec^2 \left(\dfrac{\pi x}{2}\right)} & \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\cos^2 \left(\dfrac{\pi x}{2}\right)}{x-1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{\dfrac{d}{dx}\left[\cos^2 \left(\dfrac{\pi x}{2}\right)\right]}{\dfrac{d}{dx}\left(x-1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{\pi} \; \lim\limits_{x \to 1^{-}} \; \dfrac{-2 \cos \left(\dfrac{\pi x}{2}\right) \times \sin \left(\dfrac{\pi x}{2}\right) \times \dfrac{\pi}{2}}{1} & \\ & & \\ & = \lim\limits_{x \to 1^{-}} \; - \sin \left(\dfrac{2 \pi x}{2}\right) & \left[\text{Note: }\sin 2 \theta = 2 \sin \theta \cos \theta\right] \\ & & \\ & = - \lim\limits_{x \to 1^{-}} \; \sin \left(\pi x\right) & \\ & & \\ & = 0 \end{aligned}$

Limits Indeterminate Form

If $\lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3}$ is finite, find k and the limit.

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin 2x + k \sin x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x + k \cos x}{3x^2} & \cdots (1) \end{aligned}$

Equation (1) will be of $\dfrac{0}{0}$ form if $\;\;$ $2 \cos 0 + \cos 0 = 0$

i.e. $2+k = 0$ $\implies$ $k = -2$

$\therefore$ $\;$ Equation (1) becomes

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x - 2 \cos x}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos 2x - \cos x\right)}{\dfrac{d}{dx}\left(x^2\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-2\sin 2x + \sin x}{2x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(-2\sin 2x + \sin x\right)}{\dfrac{d}{dx}\left(2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-4\cos 2x + \cos x}{2} & \\ & & \\ & = \dfrac{1}{3} \times \left(-4\cos 0 + \cos 0\right) & \\ & & \\ & = \dfrac{1}{3} \times \left(-4+1\right) = -1 \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to \frac{\pi}{4}} \; \left(\tan x\right)^{\tan 2x}$

Let $f\left(x\right)=\left(\tan x\right)^{\tan 2x}$

Then, $\log \left[f\left(x\right)\right] = \tan 2x \log \left(\tan x\right)$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} & = \lim\limits_{x \to \frac{\pi}{4}} \; \left[\tan 2x \log \left(\tan x\right)\right] & \left(\infty - 0 \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \left[\dfrac{\log \left(\tan x\right)}{\cot 2x}\right] & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}}\; \dfrac{\dfrac{d}{dx}\left[\log \left(\tan x\right)\right]}{\dfrac{d}{dx}\left(\cot 2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sec^2 x / \tan x}{-2 \text{ cosec}^2 \; 2x} & \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\dfrac{1}{\cos^2 x} \div \dfrac{\sin x}{\cos x}}{\dfrac{-2}{\sin^2 2x}} & \\ & & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{-2 \sin x \cos x} & \\ & & \\ & = - \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin^2 2x}{\sin 2x} & \left[\text{Note: }\sin 2x = 2 \sin x \cos x\right] \\ & & \\ & = - \lim\limits_{x \to \frac{\pi}{4}} \; \sin 2x & \\ & & \\ & = - \sin \left(2 \times \dfrac{\pi}{4}\right) & \\ & & \\ & = - \sin \dfrac{\pi}{2} = -1 \end{aligned}$

$\therefore$ $\;$ $\lim\limits_{x \to \frac{\pi}{4}} \; \left\{\log \left[f\left(x\right)\right] \right\} = -1$

i.e. $\log \left\{\lim\limits_{x \to \frac{\pi}{4}} \left[f\left(x\right)\right] \right\} = -1$

i.e. $\lim\limits_{x \to \frac{\pi}{4}} \left(\tan x\right)^{\tan 2x} = e^{-1} = \dfrac{1}{e}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x- \tan^{-1}x}{x^3}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\sin^{-1}x - \tan^{-1}x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin^{-1}x - \tan^{-1}x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{1+x^2}\right)}{\dfrac{d}{dx}\left(3x^2\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\dfrac{1}{2}\times \left(1-x^2\right)^{-3/2}\times \left(-2x\right)+\dfrac{2x}{\left(1+x^2\right)^2}}{6x} & \\ & & \\ & = \dfrac{1}{6} \; \lim\limits_{x \to 0} \; \left\{\dfrac{1}{\left(1-x^2\right)^{3/2}} + \dfrac{2}{\left(1+x^2\right)^2} \right\} & \\ & & \\ & = \dfrac{1}{6} \times 3 = \dfrac{1}{2} \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos x -1\right)}{\dfrac{d}{dx}\left(\cos 2x -1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\sin x}{-2\sin 2x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin x\right)}{\dfrac{d}{dx}\left(2 \sin 2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\cos x}{4 \cos 2x} & \\ & & \\ & = \dfrac{1}{4} & \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)}$

$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{4^x - 9^x}{x \left(4^x + 9^x\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(4^x - 9^x\right)}{\dfrac{d}{dx}\left[x\left(4^x + 9^x\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4^x \ln 4 - 9^x \ln 9}{4^x + 9^x + x \left(4^x \ln 4 + 9^x \ln 9\right)} \\ & & \\ & = \dfrac{\ln 4 - \ln 9}{1+1+0} \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{4}{9}\right) \\ & & \\ & = \dfrac{1}{2} \ln \left(\dfrac{2}{3}\right)^2 \\ & & \\ & = \dfrac{1}{2} \times 2 \ln \left(\dfrac{2}{3}\right) = \ln \left(\dfrac{2}{3}\right) \end{aligned}$

Limits Indeterminate Form

Evaluate $\lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)}$

$\begin{aligned} & \lim\limits_{x \to a} \; \dfrac{\log \left(x-a\right)}{\log \left(e^x - e^a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx} \left[\log \left(x-a\right)\right]}{\dfrac{d}{dx} \left[\log \left(e^x - e^a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{1}{x-a}}{\dfrac{e^x}{e^x - e^a}} \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x - e^a}{e^x \left(x-a\right)} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(e^x - e^a\right)}{\dfrac{d}{dx}\left[e^x \left(x-a\right)\right]} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{e^x}{e^x \left(x-a\right)+e^x} \\ & & \\ & = \dfrac{e^a}{e^a} = 1 \end{aligned}$

Limits Indeterminate Form

If $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$, find the value of a.

$\begin{aligned} & \lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{\dfrac{d}{dx}\left(a^x - x^a\right)}{\dfrac{d}{dx}\left(x^x - a^a\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to a} \; \dfrac{a^x \ln a - ax^{a-1}}{x^x \left(1+ \ln x\right)} & \\ & & \\ & = \dfrac{a^a \ln a -a \times a^{a-1}}{a^a \left(1+ \ln a\right)} & \\ & & \\ & = \dfrac{a^a \ln a -a^a}{a^a \left(1+ \ln a\right)} & \\ & & \\ & = \dfrac{\ln a -1}{\ln a + 1} & \end{aligned}$

$\therefore$ $\;$ $\lim\limits_{x \to a} \; \dfrac{a^x - x^a}{x^x - a^a} = -1$ $\implies$ $\dfrac{\ln a - 1}{\ln a + 1}=-1$

i.e. $\ln a -1 = -\ln a - 1$

i.e. $2 \ln a = 0$

i.e. $a = e^0 = 1$

Differentiation

Differentiate $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$ with respect to $\tan^{-1}\left(\dfrac{3x-x^3}{1-3x^2}\right)$

$\left[\begin{aligned} \text{Note: } & \tan 3\theta = \dfrac{3 \tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} \\ & \cos 2\theta = \dfrac{1-\tan^2 \theta}{1+\tan^2 \theta} \end{aligned}\right]$

Let $u = \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$, $v = \tan^{-1}\left(\dfrac{3x-x^3}{1-3x^2}\right)$

Then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} \;\; \cdots$ (1)

Let $x=\tan \alpha \;\; \cdots$ (2)

Differentiating equation(2) with respect to $\alpha$ gives

$\dfrac{dx}{d\alpha} = \sec^2 \alpha \;\; \cdots$ (3)

In view of equation (2), u becomes

$\begin{aligned} u & = \cos^{-1}\left(\dfrac{1-\tan^2 \alpha}{1+\tan^2 \alpha}\right) \\ & = \cos^{-1} \left(\cos 2 \alpha\right) \\ & = 2\alpha \end{aligned}$

$\therefore$ $\;$ $\dfrac{du}{d\alpha} = 2$ $\; \cdots$ (4)

In view of equation (2), v becomes

$\begin{aligned} v & = \tan^{-1}\left(\dfrac{3\tan \alpha - \tan^3 \alpha}{1-3\tan^2 \alpha}\right) \\ & = \tan^{-1} \left(\tan 3 \alpha\right) \\ & = 3\alpha \end{aligned}$

$\therefore$ $\;$ $\dfrac{dv}{d\alpha} = 3$ $\; \cdots$ (5)

Now, $\dfrac{du}{dx} = \dfrac{du / d\alpha}{dx / d\alpha} = \dfrac{2}{\sec^2 \alpha}$ $\;\; \cdots$ (6) [from equations (3) and (4)]

and $\dfrac{dv}{dx} = \dfrac{dv / d\alpha}{dx / d\alpha} = \dfrac{3}{\sec^2 \alpha}$ $\;\; \cdots$ (7) [from equations (3) and (5)]

$\therefore$ $\;$ In view of equations (6) and (7) equation (1) becomes

$\dfrac{du}{dv}= \dfrac{2 / \sec^2 \alpha}{3 / \sec^2 \alpha} = \dfrac{2}{3}$

Differentiation

If $f\left(x\right)=\begin{vmatrix} x & \sin x & \cos x \\ x^2 & \tan x & -x^3 \\ 2x & \sin 2x & 5x \end{vmatrix}$, $\;$ show that $\lim\limits_{x \to 0} \dfrac{f'\left(x\right)}{x}=-4$

$f'\left(x\right) = \begin{vmatrix} 1 & \sin x & \cos x \\ 2x & \tan x & -x^3 \\ 2 & \sin 2x & 5x \end{vmatrix} + \begin{vmatrix} x & \cos x & \cos x \\ x^2 & \sec^2 x & -x^3 \\ 2x & 2\cos 2x & 5x \end{vmatrix} + \begin{vmatrix} x & \sin x & -\sin x \\ x^2 & \tan x & -3x^2 \\ 2x & \sin 2x & 5 \end{vmatrix}$

$\therefore$ $\;$ $\dfrac{f'\left(x\right)}{x} = \begin{vmatrix} 1 & \sin x & \cos x \\ 2 & \tan x / x & -x^2 \\ 2 & \sin 2x & 5x \end{vmatrix} + \begin{vmatrix} 1 & \cos x & \cos x \\ x & \sec^2 x & -x^3 \\ 2 & 2 \cos 2x & 5x \end{vmatrix}+ \begin{vmatrix} 1 & \sin x & -\sin x \\ x & \tan x & -3x^2 \\ 2 & \sin 2x & 5 \end{vmatrix}$

$\begin{aligned} \therefore \; \lim\limits_{x \to 0} \dfrac{f'\left(x\right)}{x} & = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 2 & 0 & 0 \end{vmatrix} + \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 2 & 2 & 0 \end{vmatrix}+ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 5 \end{vmatrix} \\ & = -2 -2 + 0 \\ & =-4 \end{aligned}$

Differentiation

If $y = x + \tan x$, prove that $\cos^2 x \dfrac{d^2 y}{dx^2} - 2y + 2x = 0$

$ \begin{aligned} y & = x + \tan x \\ \therefore \dfrac{dy}{dx} & = 1 + \sec^2 x \\ \therefore \dfrac{d^2 y}{dx^2} & = 2 \sec x \times \sec x \tan x \\ & = 2 \sec^2 x \tan x \\ & = \dfrac{2 \tan x}{\cos^2 x} \\ \therefore \cos^2 x \dfrac{d^2 y}{dx^2} & = 2 \tan x \end{aligned} $

Since $y = x + \tan x$ $\implies$ $\tan x = y - x$

$\therefore$ $\cos^2 x \dfrac{d^2 y}{dx^2} = 2 \left(y-x\right)$

i.e. $\cos^2 x \dfrac{d^2 y}{dx^2} - 2y + 2x = 0$

Differentiation

If $y=\sec x - \tan x$, show that $\cos x \dfrac{d^2 y}{dx^2}=y^2$

$\dfrac{dy}{dx} = \sec x \tan x - \sec^2 x = \sec x \left(\tan x - \sec x\right)$
$ \begin{aligned} \therefore \dfrac{d^2 y}{dx^2} & = \sec x \left(\sec^2 x - \sec x \tan x\right) + \left(\tan x - \sec x\right) \sec x \tan x \\ & = \sec^3 x - \sec^2 x \tan x + \tan^2 x \sec x - \sec^2 x \tan x \\ & = \sec^3 x - 2 \sec^2 x \tan x + \tan^2 \sec x \\ & = \sec x \left(\sec^2 x - 2 \sec x \tan x + \tan^2 x\right) \\ & = \sec x \left(\sec x - \tan x\right)^2 \\ \therefore \dfrac{1}{\sec x} \dfrac{d^2 y}{dx^2} & = \left(\sec x - \tan x\right)^2 \\ \text{i.e. } \cos x \dfrac{d^2 y}{dx^2} & = y^2 \end{aligned} $

Differentiation

Differentiate $\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)$ with respect to x

$ \begin{aligned} \text{Let } y & = \tan^{-1} \left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right) \\ & = \tan^{-1} \left(\dfrac{1-\dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}}\right) \\ & = \tan^{-1} \left(\dfrac{1-\tan x}{1 + \tan x}\right) \end{aligned} $

Now, $\tan \left(\dfrac{\pi}{4} - x\right) = \dfrac{\tan \dfrac{\pi}{4} - \tan x}{1 + \tan \dfrac{\pi}{4} \tan x} = \dfrac{1- \tan x}{1 + \tan x}$

$\therefore$ $y = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4} - x\right)\right] = \dfrac{\pi}{4} - x$

$\therefore$ $\dfrac{dy}{dx} = -1$

Differentiation

If $x = a \sec^3 \theta$, $y = a \tan^3 \theta$, find $\dfrac{d^2 y}{dx^2}$ at $\theta = \dfrac{\pi}{3}$

$x = a \sec^3 \theta$

$\therefore$ $\dfrac{dx}{d \theta} = 3 a \sec^2 \theta \sec \theta \tan \theta = 3 a \sec^3 \theta \tan \theta$

$y = a \tan^3 \theta$

$\therefore$ $\dfrac{dy}{d \theta} = 3 a \tan^2 \theta \sec^2 \theta$

$\therefore$ $\dfrac{dy}{dx} = \dfrac{dy/d \theta}{dx / d \theta} = \dfrac{3 a \tan^2 \theta \sec^2 \theta}{3 a \sec^3 \theta \tan \theta} = \dfrac{\tan \theta}{\sec \theta}$

i.e. $\dfrac{dy}{dx} = \dfrac{\sin \theta / \cos \theta}{1 / \cos \theta} = \sin \theta$

$\therefore$ $\dfrac{d^2 y}{dx^2} = \cos \theta \times \dfrac{d \theta}{dx} = \dfrac{\cos \theta}{dx / d \theta} = \dfrac{\cos \theta}{3 a \sec^3 \theta \tan \theta}$

i.e. $\dfrac{d^2 y}{dx^2}$ $\bigg|_{\theta = \frac{\pi}{3}}$ $= \dfrac{\cos \left(\pi / 3\right)}{3a \sec^3 \left(\pi / 3\right) \tan \left(\pi / 3\right)} = \dfrac{1/2}{3a \times 2^3 \times \sqrt{3}} = \dfrac{1}{48a\sqrt{3}}$