Question

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

Answer 1

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