Determine the intervals in which $f\left(x\right) = \sin^4 x + \cos^4 x$, $x \in \left(0,\dfrac{\pi}{2}\right)$ is increasing and in which it is decreasing.
$f\left(x\right) = \sin^4 x + \cos^4 x$
$\therefore$ $f'\left(x\right) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x$
i.e. $f'\left(x\right) = 4 \sin x \cos x \left(\sin^2 x - \cos^2 x\right)$
i.e. $f'\left(x\right) = -4 \sin x \cos x \cos 2x$ $\;\; \cdots$ (1)
Now, $\sin x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$; $\cos x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$
$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$
Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) > 0$ $\implies$ $\cos 2x < 0$
i.e. $\dfrac{\pi}{2} < 2x < \pi$ or $\pi < 2x < \dfrac{3\pi}{2}$
i.e. $\dfrac{\pi}{4} < x < \dfrac{\pi}{2}$ or $\dfrac{\pi}{2} < x < \dfrac{3\pi}{4}$
Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is increasing in $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$
$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$
Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) < 0$ $\implies$ $\cos 2x > 0$
i.e. $0 < 2x < \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2} < 2x < 2\pi$
i.e. $0 < x < \dfrac{\pi}{4}$ or $\dfrac{3\pi}{4} < x < \pi $
Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is decreasing in $\left(0,\dfrac{\pi}{4}\right)$