Consider the following function y=logxcosxTaking logarithm on both sides logy=log logxcosx logy=cosx·loglogxDifferentiate both sides with respect to xUse product rule on RHS to get, ddxlogy=ddxcosx·loglogx+cosx·ddxloglogx 1ydydx=-sinx loglogx+cosx 1logxddxlogx =-sinx loglogx+cosx 1logx1x =-sinx loglogx+cosxx logx This implies that, dydx=y cosxx logx-sinx loglog xSubstitute the value of y to get, dydx=logxcosx cosxx logx-sinx loglog x

( logx )^cosx

Consider the following function . y = {(\log x)}^{cosx} Taking logarithm on both sides . logy = \log {(\log x)}^{cosx} logy = cosx \cdot \log (\log x) Differentiate both sides with respect to x Use product rule on RHS to get, \frac{d}{dx} logy = \frac{d}{dx} (cosx) \cdot \log (\log x) + cosx \cdot \frac{d}{dx} \log (\log x) \frac{1}{y} \frac{dy}{dx} = - sinx \log (\log x) + cosx \frac{1}{\log x} \frac{d}{dx} (\log x) = - sinx \log (\log x) + cosx \frac{1}{\log x} \frac{1}{x} = - sinx \log (\log x) + \frac{cosx}{x \log x} This implies that, \frac{dy}{dx} = y (\frac{cosx}{x \log x} - sinx \log (\log x)) Substitute the value of y to get, \frac{dy}{dx} = {(\log x)}^{cosx} (\frac{cosx}{x \log x} - sinx \log (\log x))

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( logx )cosx

( logx )^cosx