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

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

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

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(A) Given $y = (	an x)^{\sin x}$.Taking the natural logarithm on both sides,we get $\log y = \sin x \log(	an x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \cos x \log(	an x) + \sin x \cdot \frac{1}{	an x} \cdot \sec^2 x$.Since $\frac{\sin x}{	an x} = \cos x$,the expression simplifies to:$\frac{1}{y} \frac{dy}{dx} = \cos x \log(	an x) + \cos x \cdot \sec^2 x$.Note that $\cos x \cdot \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \sec x$.Therefore,$\frac{dy}{dx} = y \{\sec x + \cos x \log(	an x)\}$.Substituting $y = (	an x)^{\sin x}$,we get $\frac{dy}{dx} = (	an x)^{\sin x} \{\sec x + \cos x \log(	an x)\}$.

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

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