Question

solve:

$\frac{d}{dx}\left(co{s}^{-1}\sqrt{\frac{1+x}{2}}\right)$

Answer 1

Considering the question to be:

$\frac{d}{dx}\left(co{s}^{-1}\sqrt{\frac{1+x}{2}}\right)$

$=\frac{d}{du}\left({\cos }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\sqrt{\frac{1+x}{2}}\right)$

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$\frac{d}{du}\left({\cos }^{-1}\left(u\right)\right)=-\frac{1}{\sqrt{1-u^2}}$

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$\frac{d}{dx}\left(\sqrt{\frac{1+x}{2}}\right)$

$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(\frac{1+x}{2}\right)$

$=\frac{1}{2\sqrt{u}}\cdot \frac{1}{2}$

$=\frac{1}{2\sqrt{\frac{1+x}{2}}}\cdot \frac{1}{2}=\frac{1}{2\sqrt{2}\sqrt{1+x}}$

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$=(-\frac{1}{\sqrt{1-u^2}})\frac{1}{2\sqrt{2}\sqrt{1+x}}$

$=(-\frac{1}{\sqrt{1-{\left(\sqrt{\frac{1+x}{2}}\right)}^2}})\frac{1}{2\sqrt{2}\sqrt{1+x}}$

$=-\frac{1}{2\sqrt{-x+1}\sqrt{1+x}}$