Question
Differentiate the functions given in Exercise:
$(\log\text{x})^{\cos\text{x}}$

Answer

Let $\text{y}=(\log\text{x})^{\cos\text{x}}\ \dots\text{(i)}$
Taking logs on both sides, we have
$\log\text{y}=\log(\log\text{x})^{\cos\text{x}}=\cos\text{x}\log(\log\text{x})$
$\therefore\ \frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}\ [\cos\text{x}\log(\log\text{x)}]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log(\log\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}\ \ \text{[By product rule}]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log(\log\text{x})(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}.\frac{1}{\log\text{x}}-\sin\text{x}\log(\log\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos\text{x}}{\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$ $=(\log\text{x})^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$

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