Thank you for reporting, we will resolve it shortly
Q.
Find the local minimum value of the function $f\left(x\right) =sin^{4} \,x + cos^{4}\,x, 0 < x < \frac{\pi}{2}$
Application of Derivatives
Solution:
We have, $y = f\left(x\right) =sin^{4} \,x + cos^{4}\,x$
$\Rightarrow \frac{dy}{dx} = f\left(x\right) =4\,sin^{3} \,x \, cos\,x - 4\,cos^{3} \,x \, sin\,x$
$\Rightarrow \frac{dy}{dx} = - 4\,cos \,x \, sin\,x\left(cos^{2} \,x - sin^{2}\,x\right)$
$\Rightarrow \frac{dy}{dx}= -2 \,sin \,2x\, cos \,2x = -sin \,4x$
For a local maximum or a local minimum, we have
$\frac{dy}{dx}=0$
$\Rightarrow -sin\, 4x = 0$
$\Rightarrow sin\,4x = 0$
$\Rightarrow 4x = \pi$
$\Rightarrow x = \frac{\pi}{4}$
$ \left[\because 0 < x < \frac{\pi}{2} \therefore 0 < 4x < 2\pi\right]$
$\therefore \frac{d^{2}y}{dx^{2}} = -4 \,cos \,4x$
$\Rightarrow \left[\frac{d^{2}\,y}{dx^{2}}\right]_{x=\pi/4}$
So, $x = \frac{\pi}{4}$ is a point of local minimum.
The local minimum value of $f\left(x\right)$ at $x = \frac{\pi }{4}$ is
$f\left(\frac{\pi }{4}\right) = \left(sin\frac{\pi }{4}\right)^{4}+\left(cos\frac{\pi }{4}\right)^{4}$
$ = \frac{1}{2}$.