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

$\begin{array}{1 1} \large \frac{e^{sin^{-1}x}}{\sqrt(1-x^2)}\\ \large \frac{e^{sin^{-1}x}}{\sqrt(1+x^2)} \\ \large \frac{e^{cos^{-1}x}}{\sqrt(1-x^2)} \\ \large \frac{e^{cos^{-1}x}}{\sqrt(1+x^2)}\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(sinx)}{dx} $$=\large \frac{1}{\sqrt(1-x^2)}$
  • $\; \large \frac{d(e^x)}{dx} $$= e^x$
Given $y = e^{\sin^{-1}x}$
This is of the form $y = f(g(x)$, where $g(x) = \sin^{-1}x$. so, we can apply the chain rule of differentiation.
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(sinx)}{dx} $$=\large \frac{1}{\sqrt(1-x^2)}$
If $g(x) = \sin^{-1}x$, then $g'(x) = \large \frac{1}{\sqrt(1-x^2)}$
$\; \large \frac{d(e^x)}{dx} $$= e^x$
Therefore, $y' = d(e^{\sin^{-1}x})\; dx = e^{\sin^{-1}x}\; \large \frac{1}{\sqrt(1-x^2)} = \large \frac{e^{sin^{-1}x}}{\sqrt(1-x^2)}$
answered Apr 10, 2013 by balaji.thirumalai
 
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