Question

Differentiate the function $(\log x{)}^{\cos x}$ w.r.t.$x$.

Answer 1

Let $y=(\log x{)}^{\cos x}$

Taking $\log$ both sides, we get,

$\log y=\log \left(\right(\log x{)}^{\cos x})$

$\log y=\cos x.\log \left(\log x\right)$

$\left(\log \right(a^n)=n\log a)$

Differentiating both sides w.r.t. $x$,
we get

$\frac{d\left(\log y\right)}{dx}=\frac{d(\cos x.\log (\log x\left)\right)}{dx}$

$\frac{d\left(\log y\right)}{dy}\left(\frac{dy}{dx}\right)=\frac{d(\cos x.\log (\log x\left)\right)}{dx}$

$\frac{1}{y}.\frac{dy}{dx}=\frac{d(\cos x.\log (\log x\left)\right)}{dx}$

Using product rule : $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$

$\frac{1}{y}\frac{dy}{dx}=\frac{d\left(\cos x\right)}{dx}.\log \left(\log x\right)+\frac{d\left(\log \right(\log x\left)\right)}{dx}.\cos x$

$(\text{Using chain rule}\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx})$

$\frac{1}{y}\frac{dy}{dx}=-\sin x.\log \left(\log x\right)+\frac{1}{\log x}.\frac{d\left(\log x\right)}{dx}.\cos x$

$\frac{1}{y}\frac{dy}{dx}=-\sin x.\log \left(\log x\right)+\frac{\cos x}{x\log x}$

$\frac{dy}{dx}=y(-\sin x.\log (\log x)+\frac{\cos x}{x\log x})$

Substituing the value of $y=(\log x{)}^{\cos x}$,
we get

$\frac{dy}{dx}=(\log x{)}^{\cos x}(-\sin x.\log (\log x)+\frac{\cos x}{x\log x})$