Differentiate x^{sin x}, x 0 with respect t | vedclass

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Question

(A) Let $y = x^{\sin x}$. Taking the natural logarithm on both sides,we get: $\log y = \sin x \log x$ Differentiating both sides with respect to $x$ u

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Differentiate $x^{\sin x}, x > 0$ with respect to $x$.

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