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Question Number 106787 by bemath last updated on 07/Aug/20

$$\stackrel{@\mathrm{bemath}@}{}$$$$\mathrm{y}={\left(\cos 2\mathrm{x}\right)}^{\sin \mathrm{x}},\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dx}}?$$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

$$logy=sinxlog\left(cos2x\right)$$$$\frac{1}{y}\frac{dy}{dx}=sinx.\frac{-2sin2x}{cos2x}+log\left(cos2x\right)cosx$$$$\frac{dy}{dx}={\left(cos2x\right)}^{sinx}(-2sinx.tan2x+log\left(cos2x\right)cosx)$$

Answered by john santu last updated on 07/Aug/20

$$@\mathrm{JS}@$$$$\ln \left(\mathrm{y}\right)=\sin \mathrm{x}\ln \left(\cos 2\mathrm{x}\right)$$$$\frac{{\mathrm{y}}^{\prime }}{\mathrm{y}}=\cos \mathrm{x}\ln \left(\cos 2\mathrm{x}\right)+\sin \mathrm{x}\left(\frac{-2\sin 2\mathrm{x}}{\cos 2\mathrm{x}}\right)$$$${\mathrm{y}}^{\prime }=\mathrm{y}(\cos \mathrm{x}\ln \left(\cos 2\mathrm{x}\right)-2\sin \mathrm{x}\tan 2\mathrm{x})$$$$\mathrm{y}={\left(\cos 2\mathrm{x}\right)}^{\sin \mathrm{x}}\{\cos \mathrm{x}\ln (\cos$$$$2\mathrm{x})-2\sin \mathrm{x}\tan 2\mathrm{x}\}$$

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