# f: \mathbb{R}^+ \cup \{0\} \to \mathbb{R} f(x)=\arcsin(\cos \sqrt x)+\arccos(\sin \sqrt

f: $$\displaystyle{\mathbb{{{R}}}}^{+}\cup{\left\lbrace{0}\right\rbrace}\to{\mathbb{{{R}}}}$$
$$\displaystyle{f{{\left({x}\right)}}}={\arcsin{{\left({\cos{\sqrt{{x}}}}\right)}}}+{\arccos{{\left({\sin{\sqrt{{x}}}}\right)}}}$$ then what is the derivative of function f?
A)$$\displaystyle{\frac{{-{1}}}{{\sqrt{{x}}}}}$$
B) $$\displaystyle{\frac{{{1}}}{{\sqrt{{x}}}}}$$
c)$$\displaystyle{\frac{{{1}}}{{{2}\sqrt{{x}}}}}$$

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porschomcl
Applying the chain rule we get:
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{1}}}{{\sqrt{{{1}-{\left({\cos{\sqrt{{x}}}}\right)}^{{2}}}}}}}\cdot{\left(-{\sin{\sqrt{{x}}}}\right)}\cdot{\frac{{{1}}}{{{2}\sqrt{{x}}}}}-{\frac{{{1}}}{{\sqrt{{{1}-{\left({\sin{\sqrt{{x}}}}\right)}^{{2}}}}}}}\cdot{\left({\cos{\sqrt{{x}}}}\right)}\cdot{\frac{{{1}}}{{{2}\sqrt{{x}}}}}=$$
$$\displaystyle=-{\frac{{{1}}}{{{2}\sqrt{{x}}}}}\cdot{\left({\frac{{{\sin{\sqrt{{x}}}}}}{{{\left|{\sin{\sqrt{{x}}}}\right|}}}}+{\frac{{{\cos{\sqrt{{x}}}}}}{{{\left|{\cos{\sqrt{{x}}}}\right|}}}}\right)}$$
So A) is a correct answer, in the interval $$\displaystyle{\left({0},{\left({\frac{{\pi}}{{{2}}}}\right)}^{{2}}\right)}$$ where both $$\displaystyle{\sin{\sqrt{{x}}}}$$ and $$\displaystyle{\cos{\sqrt{{x}}}}$$ are positive.
###### Not exactly what you’re looking for?
Joseph Fair
Shortcut solution:
$$\displaystyle{\arcsin{{\left({\cos{{\left(\sqrt{{x}}\right)}}}\right)}}}+{\arccos{{\left({\sin{{\left(\sqrt{{x}}\right)}}}\right)}}}$$
$$\displaystyle={\arcsin{{\left({\sin{{\left({\frac{{π}}{{{2}}}}−\sqrt{{x}}\right)}}}\right)}}}+{\arccos{{\left({\cos{{\left({\frac{{\pi}}{{{2}}}}−\sqrt{{x}}\right)}}}\right)}}}$$
Using this transformation, the rest is easy.
Vasquez

Hint:
$$f(x)=\arcsin(\cos \sqrt x)+\frac{\pi}{2}−\arcsin(\sin \sqrt x)$$
$$=\frac{\pi}{2}+\arcsin(\cos \sqrt x)+\arcsin(-\sin \sqrt x)$$
Using Proof for the formula of sum of arcsine functions $$\arcsin x+\arcsin y,$$
$$f(x)=\frac{\pi}{2}+\arcsin(\cos \sqrt x |\sin \sqrt x|−\sin \sqrt x|\cos \sqrt x|)$$
as here $$x=\cos \sqrt x , \ \ \ y=−\sin \sqrt x \Rightarrow x^2+y^2=1$$
Now use $$\displaystyle{\left|{u}\right|}={\left\lbrace\begin{matrix}{u}&\ \text{ if }\ &{u}>{0}\\-{u}&&{u}\leq{0}\end{matrix}\right.}$$
For example if $$\cos \sqrt x,\sin \sqrt x$$ have the same sign $$\Leftrightarrow n\pi \leq \sqrt x \leq n\pi+\frac{\pi}{2}$$ where n is any integer