If y = (log_{x} sin x)^{x},then frac{dy}{dx} | vedclass

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(C) Given $y = (\log_{x} \sin x)^{x}$.Taking $\log$ on both sides: $\log y = x \log(\log_{x} \sin x)$.Using the change of base formula,$\log_{x} \sin

Vedclass · Accepted Answer

$y \left[ \frac{x \cot x}{\log \sin x} + \log(\log_{x} \sin x) - \frac{\log \sin x}{x (\log x)^2} \right]$

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(C) Given $y = (\log_{x} \sin x)^{x}$.Taking $\log$ on both sides: $\log y = x \log(\log_{x} \sin x)$.Using the change of base formula,$\log_{x} \sin x = \frac{\log \sin x}{\log x}$.So,$\log y = x \log \left( \frac{\log \sin x}{\log x} \right) = x [\log(\log \sin x) - \log(\log x)]$.Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 1 \cdot [\log(\log \sin x) - \log(\log x)] + x \left[ \frac{1}{\log \sin x} \cdot \frac{1}{\sin x} \cdot \cos x - \frac{1}{\log x} \cdot \frac{1}{x} \right]$.$\frac{1}{y} \frac{dy}{dx} = \log(\log_{x} \sin x) + x \left[ \frac{\cot x}{\log \sin x} - \frac{1}{x \log x} \right]$.$\frac{dy}{dx} = y \left[ \log(\log_{x} \sin x) + \frac{x \cot x}{\log \sin x} - \frac{1}{\log x} \right]$.

If $y = (\log_{x} \sin x)^{x}$,then $\frac{dy}{dx} = $

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