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

Dowqueuestbew1j 2022-01-02 Answered
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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Expert Answer

porschomcl
Answered 2022-01-03 Author has 1123 answers
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.
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Joseph Fair
Answered 2022-01-04 Author has 3439 answers
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.
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Vasquez
Answered 2022-01-08 Author has 9499 answers

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

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