frac{d}{dx}{(sin x)^x} = | vedclass

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Question

(B) Let $y = (\sin x)^x$. Taking the natural logarithm on both sides,we get $\ln y = x \ln(\sin x)$. Differentiating both sides with respect to $x$ us

Vedclass · Accepted Answer

$(\sin x)^x \left[ \frac{x\cos x + \sin x\log \sin x}{\sin x} \right]$

Vedclass · Answer

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

$\frac{d}{dx}\{(\sin x)^x\} = $

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