Question

Use logarithmic differentiation to find the derivative of given function

$f\left(x\right)=(\sin x{)}^{\cos x}$

Short answer

The derivative of given function is

$f^{\mathrm{\prime }}\left(x\right)=\left(-\sin x\ln (\sin x)+\frac{{\cos }^{2}x}{\sin x}\right)(\sin x{)}^{\cos x}$

Step-by-step solution

Step 1.Given data 

We have been given a function to find the derivative for them

$f\left(x\right)=(\sin x{)}^{\cos x}$

Step 2.Taking log on both side to get the differentiation 

$\begin{array}{r}f\left(x\right)=(\sin x{)}^{\cos x}\\ \ln f\left(x\right)=\ln (\sin x{)}^{\cos x}\text{[Taking log in both sides]}\\ \ln f\left(x\right)=\cos x\ln (\sin x)\\ \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}=\frac{d}{dx}(\cos x\ln (\sin x\left)\right)\left[\begin{array}{c}\text{Differentiate both sides}\\ \text{with respect to}x\end{array}\right]\\ \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}=\left(\frac{d}{dx}(\cos x)\right)\ln (\sin x)+\cos x\left(\frac{d}{dx}\ln (\sin x)\right)\end{array}$

$\begin{array}{r}=-\sin x\ln (\sin x)+\frac{\cos x}{\sin x}\left(\frac{d}{dx}\sin x\right)\\ =-\sin x\ln (\sin x)+\frac{\cos x}{\sin x}(\cos x)\end{array}$

Using product rule and chain rule

$f^{\mathrm{\prime }}\left(x\right)=\left(-\sin x\ln (\sin x)+\frac{\cos x}{\sin x}(\cos x)\right)f\left(x\right)$