# Derivative of $f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$

$$f:{\mathbb{R}}^{+}\cup \left\{0\right\}\to \mathbb{R}$$

$$f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$$ 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.

• Can you show us your process for finding this derivative? Jan 13, 2021 at 16:18
• Seems like you've missed a step in applying chain differentiation rule. Jan 13, 2021 at 16:20
• $f(x)=\pi-2\sqrt{x}$ Jan 13, 2021 at 16:29
• How about exploiting $\sin\sqrt{x}=\cos (\sqrt{x}-\frac\pi 2)$ and similar? Jan 13, 2021 at 16:34
• The answer A) is right, but only if $0<x<\pi^2/4$ Jan 13, 2021 at 17:13

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..

• Yes you are right I think this question is little bit awkward Jan 13, 2021 at 19:38

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.

• Thanks a lot sir Jan 13, 2021 at 19:38

Hint:

$$f(x)=\arcsin(\cos\sqrt x)+\dfrac\pi2-\arcsin(\sin\sqrt x)$$

$$=\dfrac\pi2+\arcsin(\cos\sqrt x)+\arcsin(-\sin\sqrt x)$$

$$f(x)=\dfrac\pi2+\arcsin\left(\cos\sqrt x|\sin\sqrt x|-\sin\sqrt x|\cos\sqrt x|\right)$$

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?

• Yes thanks a lot sir Jan 13, 2021 at 19:37