The derivative of (log x)^{sin x} with respec | vedclass

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(C) Let $f(x) = (\log x)^{\sin x}$ and $g(x) = \cos x$. We need to find $\frac{df}{dg} = \frac{f'(x)}{g'(x)}$. Taking the natural logarithm of $f(x)$:

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$\frac{-2}{\pi}$

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(C) Let $f(x) = (\log x)^{\sin x}$ and $g(x) = \cos x$. We need to find $\frac{df}{dg} = \frac{f'(x)}{g'(x)}$. Taking the natural logarithm of $f(x)$: $\log(f(x)) = \sin x \cdot \log(\log x)$. Differentiating with respect to $x$: $\frac{1}{f(x)} f'(x) = \cos x \cdot \log(\log x) + \sin x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$. Thus,$f'(x) = (\log x)^{\sin x} \left[ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right]$. Also,$g'(x) = -\sin x$. Therefore,$\frac{df}{dg} = \frac{(\log x)^{\sin x} [ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} ]}{-\sin x}$. At $x = \frac{\pi}{2}$,$\sin x = 1$,$\cos x = 0$,and $\log x = \log(\frac{\pi}{2})$. Substituting these values: $\frac{df}{dg} = \frac{(\log(\frac{\pi}{2}))^1 [ 0 \cdot \log(\log(\frac{\pi}{2})) + \frac{1}{(\frac{\pi}{2}) \log(\frac{\pi}{2})} ]}{-1}$. $= - \frac{1}{\frac{\pi}{2}} = -\frac{2}{\pi}$.

The derivative of $(\log x)^{\sin x}$ with respect to $\cos x$ at $x=\frac{\pi}{2}$ is

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