If y = x^{sin x},then frac{dy}{dx} = | vedclass

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(A) Given $y = x^{\sin x}$.Taking natural logarithm on both sides,we get $\ln y = \sin x \ln x$.Differentiating both sides with respect to $x$ using t

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$x^{\sin x} \left( \frac{\sin x}{x} + \cos x \ln x \right)$

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(A) Given $y = x^{\sin x}$.Taking natural logarithm on both sides,we get $\ln y = \sin x \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x}$.$\frac{1}{y} \frac{dy}{dx} = \frac{x \cos x \ln x + \sin x}{x}$.Therefore,$\frac{dy}{dx} = y \left( \frac{x \cos x \ln x + \sin x}{x} \right)$.Substituting $y = x^{\sin x}$,we get $\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x + x \cos x \ln x}{x} \right)$.

If $y = x^{\sin x}$,then $\frac{dy}{dx} = $

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