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# Differentiate the following w.r.t. x: cos x by log x

9. Differentiate the following w.r.t. x:

$\frac{\cos x }{\log x} , x > 0$

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Given function is
$f(x)=\frac{\cos x }{\log x}$
We differentiate with the help of Quotient rule
$f^{'}(x)=\frac{\frac{d(\cos x)}{dx}.\log x-\cos x.\frac{(\log x)}{dx} }{(\log x)^2 }$
$=\frac{(-\sin x).\log x-\cos x.\frac{1}{x} }{(\log x)^2 } = \frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$
Therefore, the answer is $\frac{-(x\sin x\log x+\cos x)}{x(\log x)^2}$

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