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Question

Mathematics Question on Continuity and differentiability

Differentiate the following w.r.t. xx:
sin(tan−1e−x)sin(tan^{-1}e^{-x})

Answer

The correct answer is =−e−xcos(tan−1e−x)1+e−2x=\frac{-e^{-x}cos(tan^{-1}e^{-x})}{1+e^{-2x}}
Let y=sin(tan−1e−x)y=sin(tan^{-1}e^{-x})
By using the chain rule,we obtain
dydx=ddx[sin(tan−1e−x)]\frac{dy}{dx}=\frac{d}{dx}[sin(tan^{-1}e^{-x})]
=cos(tan−1e−x).ddx(tan−1e−x)=cos(tan^{-1}e^{-x}).\frac{d}{dx}(tan^{-1}e^{-x})
=cos(tan−1e−x).11+(e−x)2.ddx(e−x)=cos(tan^{-1}e^{-x}).\frac{1}{1+(e^{-x})^2}.\frac{d}{dx}(e^{-x})
=cos(tan−1e−x)1+e−2x.e−x.ddx(−x)=\frac{cos(tan^{-1}e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{d}{dx}(-x)
=e−xcos(tan−1e−x)1+e−2x×(−1)=\frac{e^{-x}cos(tan^{-1}e^{-x})}{1+e^{-2x}}\times(-1)
=−e−xcos(tan−1e−x)1+e−2x=\frac{-e^{-x}cos(tan^{-1}e^{-x})}{1+e^{-2x}}