If y = (tan x)^{cot x},then frac{dy}{dx} = | vedclass

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(A) Given $y = (	an x)^{\cot x}$.Taking natural logarithm on both sides,we get $\log y = \cot x \log(	an x)$.Differentiating both sides with respect

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$y \csc^2 x (1 - \log 	an x)$

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(A) Given $y = (	an x)^{\cot x}$.Taking natural logarithm on both sides,we get $\log y = \cot x \log(	an x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cot x) \cdot \log(	an x) + \cot x \cdot \frac{d}{dx}(\log(	an x))$.Since $\frac{d}{dx}(\cot x) = -\csc^2 x$ and $\frac{d}{dx}(\log(	an x)) = \frac{1}{	an x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin(2x)} = 2 \csc(2x)$ or simply $\csc^2 x \cdot \cot x$ is not correct,let us re-evaluate: $\frac{d}{dx}(\log(	an x)) = \frac{\sec^2 x}{	an x} = \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x} = \csc x \sec x$.Wait,$\frac{d}{dx}(\log(	an x)) = \frac{1}{	an x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.Actually,$\frac{d}{dx}(\log(	an x)) = \frac{\sec^2 x}{	an x} = \frac{1}{\sin x \cos x} = \csc x \sec x$.Let us re-calculate: $\frac{1}{y} \frac{dy}{dx} = -\csc^2 x \log(	an x) + \cot x \cdot (\frac{\sec^2 x}{	an x}) = -\csc^2 x \log(	an x) + \cot x \cdot \frac{1}{\sin x \cos x} \cdot \cos x = -\csc^2 x \log(	an x) + \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x \cos x} = -\csc^2 x \log(	an x) + \csc^2 x = \csc^2 x (1 - \log 	an x)$.Therefore,$\frac{dy}{dx} = y \csc^2 x (1 - \log 	an x)$.

If $y = (\tan x)^{\cot x}$,then $\frac{dy}{dx} =$

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