Consider the following function y=sinxcosxTaking logarithm on both sides logy=log sinxcosx logy=cosx·logsinxDifferentiate both sides with respect to xUse product rule on RHS to get, ddxlogy=ddxcosx·logsinx+cosx·ddxlogsinx 1ydydx=-sinx logsinx+cosx 1sinxddxsinx =-sinx logsinx+cosx cosxsinx =-sinx logsinx+cosx cot xThis implies that, dydx=y cosx cot x-sinx logsinxSubstitute the value of y to get, dydx=sinxcosx cosx cot x-sinx logsinx

(sinx)^cosx

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

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(sinx)cosx

(sinx)^cosx