Differentiate the function with respect to x: | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Differentiate the function with respect to $x$: $(\log x)^{\cos x}$

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