Find the intervals in which the function f(x) = sin^4x + cos^4x is increasing or decreasing where the range of x is ( 0, pie/2) .

the given function is $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$
differentiating wrt x:
$$\begin{gathered} f\text{'}\left(x\right)=4{\sin }^{3}x\cos x+4{\cos }^{3}x(-\sin x) \\ =4\sin x\cos x({\sin }^{2}x-{\cos }^{2}x) \\ =-2\left(2\sin x\cos x\right)\cos 2x \\ =-2\sin 2x\cos 2x \\ f\text{'}\left(x\right)=-\sin 4x \end{gathered}$$
we have : $x\in [0,\frac{\pi }{2}]\Rightarrow 4x\in [0,2\mathrm{\pi }]$
$$\begin{gathered} \sin 4x⩾0 for x\in [0,\frac{\pi }{4}] and \\ \sin 4x⩽0 for x\in [\frac{\pi }{4},\frac{\pi }{2}] \end{gathered}$$$$\begin{gathered} \sin 4x⩾0 for x\in [0,\frac{\pi }{4}] and \\ \sin 4x⩽0 for x\in [\frac{\pi }{4},\frac{\pi }{2}] \end{gathered}$$
therefore
$f\text{'}\left(x\right)<0 for 0<x<\frac{\pi }{4}$
so f(x) is decreasing on $[0,\frac{\mathrm{\pi }}{4}]$
f'(x)>0 for $\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2}$
so f(x) is increasing on $[\frac{\pi }{4},\frac{\pi }{2}]$

hope this helps you