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Differentiate the following w.r.t. \(x : e^x + e^{x^2} + \: ... + e^{x^5} \)

$\begin{array}{1 1} e^x +2\;x e^{x^2} + 3\;x^2e^{x^3} + 4\;x^3e^{x^4}+ 5\;x^4e^{x^5} \\ e^x + x e^{x^2} + 2\;x^2e^{x^3} + 3\;x^3e^{x^4}+ 4\;x^4e^{x^5} \\ e^x +2 e^{x^2} + 3 e^{x^3} + 4e^{x^4}+ 5e^{x^5} \\ e^x -2\;x e^{x^2} + 3\;x^2e^{x^3} - 4\;x^3e^{x^4}+ 5\;x^4e^{x^5} \end{array} $

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Toolbox:
  • According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
  • Using chain rule, it can be shown that $\; \large \frac{d(e^{x^n})}{dx} $$= n\;x^{n-1}e^{x^n}$
Given $y = e^x + e^{x^2} + \: ... + e^{x^5} = e^x + e^{x^2} + e^{x^3} + e^{x^4}+ e^{x^5}$
Using chain rule, it can be shown that $\; \large \frac{d(e^{x^n})}{dx} $$= n\;x^{n-1}e^{x^n}$
$\Rightarrow y' = e^x +2\;x^{2-1} e^{x^2} + 3\;x^{3-1}e^{x^3} +4\;x^{4-1} e^{x^4}+ 5\;x^{5-1}e^{x^5}$
$\Rightarrow y' = e^x +2\;x e^{x^2} + 3\;x^2e^{x^3} + 4\;x^3e^{x^4}+ 5\;x^4e^{x^5}$
answered Apr 10, 2013 by balaji.thirumalai
 
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