The derivative of (sin x)^x with respect to x | vedclass

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(A) Let $u = (\sin x)^x$ and $v = x^{\sin x}$.Taking log on both sides for $u$:$\log u = x \log (\sin x)$.Differentiating with respect to $x$:$\frac{1

Vedclass · Accepted Answer

$\frac{(\sin x)^{x-1}[\sin x \log (\sin x)+x \cos x]}{x^{\sin x-1}[\sin x+x \cos x \log x]}$

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(A) Let $u = (\sin x)^x$ and $v = x^{\sin x}$.Taking log on both sides for $u$:$\log u = x \log (\sin x)$.Differentiating with respect to $x$:$\frac{1}{u} \frac{du}{dx} = \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log (\sin x) + x \cot x$.$\frac{du}{dx} = (\sin x)^x [\log (\sin x) + x \cot x] = (\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]$.Taking log on both sides for $v$:$\log v = \sin x \log x$.Differentiating with respect to $x$:$\frac{1}{v} \frac{dv}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{x \cos x \log x + \sin x}{x}$.$\frac{dv}{dx} = x^{\sin x} \cdot \frac{x \cos x \log x + \sin x}{x} = x^{\sin x-1} [\sin x + x \cos x \log x]$.Therefore, $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]}{x^{\sin x-1} [\sin x + x \cos x \log x]}$.

The derivative of $(\sin x)^x$ with respect to $x^{(\sin x)}$ is

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