Question

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

• A. $\left(\frac{1}{2}\right)\sqrt{(secx+1)}$
• B. $\sqrt{(secx+1)}$
• C. $-\left(\frac{1}{2}\right)\sqrt{(secx+1)}$
• D. $-\sqrt{(secx+1)}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the value$\frac{d}{dx}{\cos }^{-1}\sqrt{\cos x}$:

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

Given,

$$\begin{gathered} \frac{d}{dx}{\cos }^{-1}\left(\sqrt{\cos \left(x\right)}\right)=\frac{-1}{\sqrt{1-\cos x}}\times \frac{d}{dx}\sqrt{\cos \left(x\right)} \\ =\frac{-1}{\sqrt{1-\cos x}}\times \frac{1}{2\sqrt{\cos \left(x\right)}}\times \frac{d}{dx}\left(\cos \left(x\right)\right) \\ =\frac{-1}{\sqrt{1-\cos x}}\times \frac{1}{2\sqrt{\cos \left(x\right)}}\times -\sin \left(x\right) \\ =\frac{\sin \left(x\right)}{2\sqrt{1-\cos x}\sqrt{\cos \left(x\right)}} \end{gathered}$$

Since ${\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{1}$

$\frac{d}{dx}{\cos }^{-1}\left(\sqrt{\cos \left(x\right)}\right)=\frac{\sqrt{1-\cos x}\sqrt{1+\cos \left(x\right)}}{2\sqrt{1-\cos x}\sqrt{\cos \left(x\right)}}\mathbf{}\left[\because \sin \left(x\right)=\sqrt{1-{\cos }^{2}x}=\sqrt{1-\mathrm{cosx}}\sqrt{1+\mathrm{cosx}}\right]$

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

Hence, option(A) is correct.