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

$\begin{array}{1 1} - \large\frac{cos(tan^{-1}e^{-x})e^{-x}}{1+e^{-2x}} \\ - \large\frac{sin(tan^{-1}e^{-x})e^{-x}}{1+e^{-2x}} \\ - \large\frac{cos(tan^{-1}e^{-x})e^{-x}}{1-2e^{-2x}} \\ - \large\frac{sin(tan^{-1}e^{-x})e^{-x}}{1-2e^{-2x}} \end{array}$

## 1 Answer

Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(e^{-x})}{dx} $$= -e^{-x} • \; \large \frac{d(tan^{-1}x)}{dx}$$= \large\frac{1}{1+x^2}$
• $\; \large \frac{d(sinx)}{dx} $$= cosx Let y =sin\;( tan^{-1} e^{-x}) This is of the form y = f(g(x), where g(x) = \tan^{-1}e^{-x}. so, we can apply the chain rule of differentiation. Now, g(x) itself is of the form g(x) = f_1(f_2(x)), where f_2 = e^{-x}, so let us apply the chain rule and calculate g'(x) According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \; \large \frac{d(e^{-x})}{dx}$$= -e^{-x}$
$\; \large \frac{d(tan^{-1}x)}{dx} $$= \large\frac{1}{1+x^2} \Rightarrow g'(x) = \large\frac{1}{1+(e^{-x})^2}$$\;-e^{-x} = - \large\frac{e^{-x}}{1+e^{-2x}}$
$\; \large \frac{d(sinx)}{dx}$$= cosx$
$\Rightarrow y' = cos(tan^{-}e^{-x})\times - \large\frac{e^{-x}}{1+e^{-2x}}$
$\Rightarrow y' = - \large\frac{cos(tan^{-1}e^{-x})e^{-x}}{1+e^{-2x}}$
answered Apr 10, 2013

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