frac{d}{dx}(log tan x) = | vedclass

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Question

(B) Let $y = \log(	an x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{	an x} \cdot \frac{d}{dx}(	an x)$$\fr

Vedclass · Accepted Answer

$2\csc 2x$

Vedclass · Answer

(B) Let $y = \log(	an x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{	an x} \cdot \frac{d}{dx}(	an x)$$\frac{dy}{dx} = \frac{1}{	an x} \cdot \sec^2 x$Since $	an x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we have:$\frac{dy}{dx} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x}$Multiplying the numerator and denominator by $2$:$\frac{dy}{dx} = \frac{2}{2 \sin x \cos x}$Using the identity $\sin 2x = 2 \sin x \cos x$:$\frac{dy}{dx} = \frac{2}{\sin 2x} = 2 \csc 2x$.Thus,the correct option is $B$.

$\frac{d}{dx}(\log \tan x) = $

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