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Question

# $\frac{d}{dx}{\mathrm{cos}}^{-1}\sqrt{\mathrm{cos}x}$

A

$\left(\frac{1}{2}\right)\sqrt{\left(secx+1\right)}$

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B

$\sqrt{\left(secx+1\right)}$

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C

$-\left(\frac{1}{2}\right)\sqrt{\left(secx+1\right)}$

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D

$-\sqrt{\left(secx+1\right)}$

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Solution

## The correct option is A $\left(\frac{1}{2}\right)\sqrt{\left(secx+1\right)}$Explanation for the correct option:Find the value$\frac{d}{dx}{\mathrm{cos}}^{-1}\sqrt{\mathrm{cos}x}$: $\frac{d}{dx}{\mathrm{cos}}^{-1}\left(x\right)=\frac{-1}{\sqrt{1-{x}^{2}}}$Given, $\frac{d}{dx}{\mathrm{cos}}^{-1}\left(\sqrt{\mathrm{cos}\left(x\right)}\right)=\frac{-1}{\sqrt{1-\mathrm{cos}x}}×\frac{d}{dx}\sqrt{\mathrm{cos}\left(x\right)}\phantom{\rule{0ex}{0ex}}=\frac{-1}{\sqrt{1-\mathrm{cos}x}}×\frac{1}{2\sqrt{\mathrm{cos}\left(x\right)}}×\frac{d}{dx}\left(\mathrm{cos}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{-1}{\sqrt{1-\mathrm{cos}x}}×\frac{1}{2\sqrt{\mathrm{cos}\left(x\right)}}×-\mathrm{sin}\left(x\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(x\right)}{2\sqrt{1-\mathrm{cos}x}\sqrt{\mathrm{cos}\left(x\right)}}$Since ${\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{1}$$\frac{d}{dx}{\mathrm{cos}}^{-1}\left(\sqrt{\mathrm{cos}\left(x\right)}\right)=\frac{\sqrt{1-\mathrm{cos}x}\sqrt{1+\mathrm{cos}\left(x\right)}}{2\sqrt{1-\mathrm{cos}x}\sqrt{\mathrm{cos}\left(x\right)}}\mathbf{}\left[\because \mathrm{sin}\left(x\right)=\sqrt{1-{\mathrm{cos}}^{2}x}=\sqrt{1-\mathrm{cosx}}\sqrt{1+\mathrm{cosx}}\right]$ $\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\sqrt{\mathbf{secx}\mathbf{+}\mathbf{1}}\mathbf{\right)}$Hence, option(A) is correct.

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