Dowqueuestbew1j

2022-01-02

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

porschomcl

Applying the chain rule we get:
${f}^{\prime }\left(x\right)=\frac{1}{\sqrt{1-{\left(\mathrm{cos}\sqrt{x}\right)}^{2}}}\cdot \left(-\mathrm{sin}\sqrt{x}\right)\cdot \frac{1}{2\sqrt{x}}-\frac{1}{\sqrt{1-{\left(\mathrm{sin}\sqrt{x}\right)}^{2}}}\cdot \left(\mathrm{cos}\sqrt{x}\right)\cdot \frac{1}{2\sqrt{x}}=$
$=-\frac{1}{2\sqrt{x}}\cdot \left(\frac{\mathrm{sin}\sqrt{x}}{|\mathrm{sin}\sqrt{x}|}+\frac{\mathrm{cos}\sqrt{x}}{|\mathrm{cos}\sqrt{x}|}\right)$
So A) is a correct answer, in the interval $\left(0,{\left(\frac{\pi }{2}\right)}^{2}\right)$ where both $\mathrm{sin}\sqrt{x}$ and $\mathrm{cos}\sqrt{x}$ are positive.

Joseph Fair

Shortcut solution:
$\mathrm{arcsin}\left(\mathrm{cos}\left(\sqrt{x}\right)\right)+\mathrm{arccos}\left(\mathrm{sin}\left(\sqrt{x}\right)\right)$
$=\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{\pi }{2}-\sqrt{x}\right)\right)+\mathrm{arccos}\left(\mathrm{cos}\left(\frac{\pi }{2}-\sqrt{x}\right)\right)$
Using this transformation, the rest is easy.

Vasquez

Hint:
$f\left(x\right)=\mathrm{arcsin}\left(\mathrm{cos}\sqrt{x}\right)+\frac{\pi }{2}-\mathrm{arcsin}\left(\mathrm{sin}\sqrt{x}\right)$
$=\frac{\pi }{2}+\mathrm{arcsin}\left(\mathrm{cos}\sqrt{x}\right)+\mathrm{arcsin}\left(-\mathrm{sin}\sqrt{x}\right)$
Using Proof for the formula of sum of arcsine functions $\mathrm{arcsin}x+\mathrm{arcsin}y,$
$f\left(x\right)=\frac{\pi }{2}+\mathrm{arcsin}\left(\mathrm{cos}\sqrt{x}|\mathrm{sin}\sqrt{x}|-\mathrm{sin}\sqrt{x}|\mathrm{cos}\sqrt{x}|\right)$
as here
Now use
For example if $\mathrm{cos}\sqrt{x},\mathrm{sin}\sqrt{x}$ have the same sign $⇔n\pi \le \sqrt{x}\le n\pi +\frac{\pi }{2}$ where n is any integer

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