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.