Differentiate the following w.r.t. \(x\):
\(sin(tan^{-1}e^{-x})\)
\(sin(tan^{-1}e^{-x})\)
Solution and Explanation
Let \(y=sin(tan^{-1}e^{-x})\)
By using the chain rule,we obtain
\(\frac{dy}{dx}=\frac{d}{dx}[sin(tan^{-1}e^{-x})]\)
\(=cos(tan^{-1}e^{-x}).\frac{d}{dx}(tan^{-1}e^{-x})\)
\(=cos(tan^{-1}e^{-x}).\frac{1}{1+(e^{-x})^2}.\frac{d}{dx}(e^{-x})\)
\(=\frac{cos(tan^{-1}e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{d}{dx}(-x)\)
\(=\frac{e^{-x}cos(tan^{-1}e^{-x})}{1+e^{-2x}}\times(-1)\)
\(=\frac{-e^{-x}cos(tan^{-1}e^{-x})}{1+e^{-2x}}\)
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.