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Differentiate the following w.r.t. \(x : \sqrt {e^\sqrt x}, x > 0 \)

$\begin{array}{1 1} \large \frac{1}{4} \frac{e^\sqrt x}{\sqrt (x\; e^\sqrt x)} \\ \large \frac{1}{2} \frac{1}{ (x\; e^\sqrt x)} \\ \large \frac{1}{4} \frac{e^\sqrt x}{(x\; e^\sqrt x)} \\ \large \frac{1}{2} \frac{1}{\sqrt (x\; e^\sqrt x)}\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} $$= n\;x^{n-1}e^{x^n}$
Given $y =x : \sqrt {e^\sqrt x},\; x > 0$
We can rewrite $y = (e^\sqrt x)^{\large\frac{1}{2}}$
This is of the form $y = f(g(x))$, where $g(x) = e^ \sqrt x$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
Using chain rule, it can be shown that $\; \large \frac{d(e^{x^n})}{dx} $$= n\;x^{n-1}e^{x^n}$
$\Rightarrow g'(x) = \frac{1}{2}\; x^{\frac{1}{2} -1} e^\sqrt x = \large \frac{e^\sqrt x}{2\sqrt x}$
$\Rightarrow y' = \large \frac{1}{2}$$ (e^\sqrt x)^{\frac{1}{2}-1} \times \large \frac{e^\sqrt x}{2\sqrt x}$
$\Rightarrow y' = \large \frac{1}{4} \frac{e^\sqrt x}{\sqrt (x\; e^\sqrt x)}$
answered Apr 10, 2013 by balaji.thirumalai
 
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