$f:{\mathbb{R}}^{+}\cup \left\{0\right\}\to \mathbb{R}$
$f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$ then what is the derivative of function $f?$
A)$\frac{-1}{\sqrt{x}}$
B)$\frac{1}{\sqrt{x}}$
c)$\frac{1}{2\sqrt{x}}$
Here is my solution: I found derivative of function is $$\frac{-1}{2\sqrt{x}}\left(\frac{1}{\sin\sqrt{x}}+\frac{1}{\cos\sqrt{x}}\right).$$
But I can not reach the answer that given in question. Any help will be appreciated.
Applying the chain rule we get: $\\ f'(x)=\frac{1}{\sqrt{1-(\cos\sqrt{x})^2}}\cdot (-\sin\sqrt{x})\cdot\frac{1}{2\cdot\sqrt{x}}-\frac{1}{\sqrt{1-(\sin\sqrt{x})^2}}\cdot (\cos\sqrt{x})\cdot\frac{1}{2\cdot\sqrt{x}}=\\=-\frac{1}{2\cdot\sqrt{x}}\cdot(\frac{\sin\sqrt{x}}{|\sin\sqrt{x}|}+\frac{\cos\sqrt{x}}{|\cos\sqrt{x}|})$
So the correct answer, as @Bernard Massé was saying, is the first one, but only in the interval $(0,(\frac{\pi}{2})^2)$, where both $\sin\sqrt{x}$ and $\cos\sqrt{x}$ are positive..
Hint:
$$f(x)=\arcsin(\cos\sqrt x)+\dfrac\pi2-\arcsin(\sin\sqrt x)$$
$$=\dfrac\pi2+\arcsin(\cos\sqrt x)+\arcsin(-\sin\sqrt x)$$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$f(x)=\dfrac\pi2+\arcsin\left(\cos\sqrt x|\sin\sqrt x|-\sin\sqrt x|\cos\sqrt x|\right) $$
as here $x=\cos\sqrt x,y=-\sin\sqrt x\implies x^2+y^2=1$
Now use $|u|=\begin{cases}u&\mbox{if } u>0 \\-u &u\le0\end{cases}$
For example if $\cos\sqrt x,\sin\sqrt x$ have the same sign $\iff n\pi\le\sqrt x\le n\pi+\dfrac\pi2$ where $n$ is any integer
$$f(x)=\dfrac\pi2+\arcsin0=?$$
Can you take it from here?
Shortcut solution:
$$\arcsin\left(\cos(\sqrt{x})\right)+\arccos\left(\sin(\sqrt{x})\right) \\=\arcsin\left(\sin\left(\frac\pi2-\sqrt{x}\right)\right)+\arccos\left(\cos\left(\frac\pi2-\sqrt{x}\right)\right).$$
Using this transformation, the rest is easy.