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

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

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$(\sin x)^{	an x}(1+\sec^2 x \log(\sin x))$

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(A) Given $y=(\sin x)^{	an x}$.Taking the natural logarithm on both sides,we get $\log y = 	an x \cdot \log(\sin x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(	an x) \cdot \log(\sin x) + 	an x \cdot \frac{d}{dx}(\log(\sin x))$.$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + 	an x \cdot \frac{1}{\sin x} \cdot \cos x$.Since $\frac{\cos x}{\sin x} = \cot x$,we have $	an x \cdot \cot x = 1$.Thus,$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1$.Multiplying by $y$,we get $\frac{dy}{dx} = y[1 + \sec^2 x \log(\sin x)]$.Substituting $y = (\sin x)^{	an x}$,we get $\frac{dy}{dx} = (\sin x)^{	an x}[1 + \sec^2 x \log(\sin x)]$.

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

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