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Differentiate the following w.r.t. $x: \large \frac {\cos x}{\log x}$,$\; x > 0 $

$\begin{array}{1 1} - \large \frac{x\; \sin x \log x + \cos x}{x\;(\log x)^2} \\ \large \frac{x\; \sin x \log x + \cos x}{x\;(\log x)^2} \\ - \large \frac{x\; \sin x \log x - \cos x}{x\;\log x} \\ - \large \frac{x\; \sin x \log x + \cos x}{x\;\log x}\end{array} $

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Toolbox:
  • According to the Quotient Rule for differentiation, given two functions $u$ and $v,\; \large (\frac{u}{v})\;$$' = \large \frac{1}{v^2}$ $(v\;u' - u\;v')$
  • $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x}$
  • $\; \large \frac{d(cosx)}{dx} $$=-sinx $
Given $y = \large \frac {\cos x}{\log x}$.
This is of the form $\large \frac{u}{v}$ where $u = \cos x$ and $v = \log x$
If $u = \cos x$, $u' = -\sin x$
If $u = \log x$, $u' = \large \frac{1}{x}$
According to the Quotient Rule for differentiation, given two functions $u$ and $v,\; \large (\frac{u}{v})\;$$' = \large \frac{1}{v^2}$ $(v\;u' - u\;v')$
$\Rightarrow y' = \large \frac{1}{(\log x)^2}$$(\log x\; (-\sin x) - \cos x (\large \frac{1}{x}$)$)$
$\Rightarrow y' = - \large \frac{x\; \sin x \log x + \cos x}{x\;(\log x)^2}$
answered Apr 4, 2013 by thanvigandhi_1
edited Apr 10, 2013 by balaji.thirumalai
 
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