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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} $

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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 by balaji.thirumalai
 
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