I know that it can be obtained by simply differentiating the equation and finding the roots of the derivative, but it is a lengthy and tricky process. I am looking for a faster and more straightforward way.
A more effective and quick way to find the answer via simple differentiation will also be appreciated.
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Hint: Use the fact that $\arcsin x+\arccos x=\frac{\pi}{2}$ on $x\in[-1,1]$: let $t=\arccos x\in[0,\pi]$, $\arcsin x=\frac{\pi}{2}-t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and you need to minimise/maximise $(\frac{\pi}{2}-t)^4+t^4$. Now it is a lot easier to find the derivatives etc.
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Let $y=\arcsin x$, so you're trying to extremise $y^4+(\frac{\pi}{2}-y)^4$. The first two derivatives with respect to $y$ are proportional to $y^3-(\frac{\pi}{2}-y)^3,\,y^2+(\frac{\pi}{2}-y)^2$. The former vanishes when $y=\frac{\pi}{4}$ i.e. $x=\frac{1}{\sqrt{2}}$, while the latter is non-negative so we have obtained a minimum of the original function. In particular, $\arcsin^4\frac{1}{\sqrt{2}}+\arccos^4\frac{1}{\sqrt{2}}=\frac{\pi^4}{8}$. To find the maximum, go to the extreme values of $x$ in the original function, namely $\pm 1$. You'll find the maximum is at $x=-1$, giving $(-\frac{\pi}{2})^4+\pi^4=\frac{17\pi^4}{16}$.
Let $a = \arcsin x , b = \arccos x$, then you need to optimize $a^4+b^4$ with the constraint $a+b =\dfrac{\pi}{2}.$ An obvious lower bound follows from power-mean inequality: $$a^4+b^4\geq2\left(\dfrac{a+b}{2}\right)^4 = \dfrac{\pi^4}{8}.$$
In order to maximize though, you will need to resort to Lagrange multipliers, or second derivative test for single variables by eliminating one of them.