If y = tan(cos^{-1} x), then frac{dy}{dx} is | vedclass

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(D) Given $y = 	an(\cos^{-1} x)$.Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$.We know that $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos

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$\frac{-1}{x^2 \sqrt{1-x^2}}$

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(D) Given $y = 	an(\cos^{-1} x)$.Let $\cos^{-1} x = 	heta$, then $\cos 	heta = x$.We know that $	an 	heta = \frac{\sqrt{1 - \cos^2 	heta}}{\cos 	heta} = \frac{\sqrt{1 - x^2}}{x}$.So, $y = \frac{\sqrt{1 - x^2}}{x}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1 - x^2}) - \sqrt{1 - x^2} \cdot \frac{d}{dx}(x)}{x^2}$$\frac{dy}{dx} = \frac{x \cdot \frac{-2x}{2\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$$\frac{dy}{dx} = \frac{\frac{-x^2}{\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$$\frac{dy}{dx} = \frac{-x^2 - (1 - x^2)}{x^2 \sqrt{1 - x^2}} = \frac{-1}{x^2 \sqrt{1 - x^2}}$.

If $y = \tan(\cos^{-1} x)$, then $\frac{dy}{dx}$ is equal to

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