Find the derivative of cos^{-1}sqrt{cos x} wi | vedclass

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Question

(A) Let $y = \cos^{-1}\sqrt{\cos x}$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{\cos x})^2}} \cdot \frac{d}{dx}(\sqrt{\cos x})$.

Vedclass · Accepted Answer

$\frac{1}{2}\sqrt{1 + \sec x}$

Vedclass · Answer

(A) Let $y = \cos^{-1}\sqrt{\cos x}$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{\cos x})^2}} \cdot \frac{d}{dx}(\sqrt{\cos x})$.$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - \cos x}} \cdot \frac{1}{2\sqrt{\cos x}} \cdot (-\sin x)$.$\frac{dy}{dx} = \frac{\sin x}{2\sqrt{\cos x}\sqrt{1 - \cos x}}$.Since $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{(1 - \cos x)(1 + \cos x)}$,we have:$\frac{dy}{dx} = \frac{\sqrt{1 - \cos x}\sqrt{1 + \cos x}}{2\sqrt{\cos x}\sqrt{1 - \cos x}}$.$\frac{dy}{dx} = \frac{\sqrt{1 + \cos x}}{2\sqrt{\cos x}} = \frac{1}{2}\sqrt{\frac{1 + \cos x}{\cos x}} = \frac{1}{2}\sqrt{\sec x + 1}$.

Find the derivative of $\cos^{-1}\sqrt{\cos x}$ with respect to $x$.

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