Question
Differentiate the function ${\left( {\log x} \right)^{\cos x}}$ w.r.t. x.
Taking log on both sides, we have
$\Rightarrow \log y = \log {\left( {\log x} \right)^{\cos x}} = \cos x\log \left( {\log x} \right)$
$\Rightarrow \frac{d}{{dx}}\log y = \frac{d}{{dx}}\left[ {\cos x\log \left( {\log x} \right)} \right]$
$\Rightarrow \frac{1}{y}\frac{dy}{{dx}} = \cos x\frac{d}{{dx}}\log \left( {\log x} \right) + \log \left( {\log x} \right)\frac{d}{{dx}}\cos x$ [By Product rule]
$ \Rightarrow \frac{1}{y}.\frac{{dy}}{{dx}} = \cos x\frac{1}{{\log x}}\frac{d}{{dx}}\left( {\log x} \right) + \log \left( {\log x} \right)\left( { - \sin x} \right)$
$\Rightarrow \frac{1}{y}.\frac{{dy}}{{dx}} = \frac{{\cos x}}{{\log x}}.\frac{1}{x} - \sin x\log \left( {\log x} \right)$
$\Rightarrow \frac{{dy}}{{dx}} = y\left[ {\frac{{\cos x}}{{x\log x}} - \sin x\log \left( {\log x} \right)} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = {\left( {\log x} \right)^{\cos x}}\left[ {\frac{{\cos x}}{{x\log x}} - \sin x\log \left( {\log x} \right)} \right]$
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