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Differentiate the following w.r.t. \(x : e^{x^{3} } \)

$\begin{array}{1 1} 3x^2\;e^{x^3} \\3x\;e^{x^3} \\ 3x^{-2}\;e^{x^2} \\ 3x^2\;e^{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(e^x)}{dx} $$= e^x$
Given $y = e^{x^3}$
This is of the form $y = f(g(x))$, where $g(x) = x^3$. We can solve this using the chain rule of differentiation.
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
If $g(x) = x^3$, then $g'(x) = 3\;x^2$
$\; \large \frac{d(e^x)}{dx} $$= e^x$
$\Rightarrow y' = e^{x^3}\; 3\;x^2 = 3x^2\;e^{x^3}$
answered Apr 10, 2013 by balaji.thirumalai
 
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