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Differentiate the following w.r.t. $ x: $ $ cos \: ( log\: x + e^x ), x > 0 $

$\begin{array}{1 1} -sin (\log x + e^x)\; (\large \frac{1}{x}+ e^x) \\sin (\log x + e^x)\; (\large \frac{1}{x}- e^x) \\ sin (\log x + e^x)\; (\large \frac{1}{x}+ e^x) \\-sin (\log x + e^x)\; (\large \frac{1}{x}- e^x) \end{array} $

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  • 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(cosx)}{dx} $$=-sinx $
  • $\; \large \frac{d(e^{x})}{dx} $$= e^{x}$
  • $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x}$
Given $y = cos \: ( log\: x + e^x ), x > 0$
This is of the form $y = f(g(x)$, where $g(x) =\log x + e^x ,\; x > 0$, so we can apply the chain rule of differentiation.
We know that $\; \large \frac{d(e^{x})}{dx} $$= e^{x}$ and $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x}$
$\Rightarrow g'(x) = \large \frac{1}{x}$$+ e^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(cosx)}{dx} $$=-sinx $
$\Rightarrow y' = -sin (\log x + e^x)\; (\large \frac{1}{x}$$+ e^x)$
answered Apr 4, 2013 by thanvigandhi_1
edited Apr 10, 2013 by balaji.thirumalai
 
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