Chapter 4.4, Problem 43E

To determine

To find $\frac{dy}{dx}$ of the given function by using logarithmic differentiation.

Answer to Problem 43E

  $\frac{dy}{dx}$ of the function $y={\left(\sin x\right)}^x,0<x<\frac{\pi }{2}$ is $\frac{dy}{dx}={\sin }^x\left(x\right)\left(\left(\ln \left(\sin x\right)\right)+x\cot x\right)$ when $x=0$ it is undefined and when $x=\frac{\pi }{2}$ it is $0$ .

Explanation of Solution

Given information:

  $y={\left(\sin x\right)}^x,0<x<\frac{\pi }{2}$

Calculation:

  $\begin{array}{c}\ln y=\ln {\left(\sin x\right)}^x\\ \ln y=x\ln \left(\sin x\right)\\ \frac{d}{dx}\left(\ln y\right)=\frac{d}{dx}\left(x\sin x\right)\\ \frac{1}{y}\frac{dy}{dx}=x\cot x+\ln \sin x\\ \frac{dy}{dx}=y\left(x\cot x+1\ln \sin x\right)\\ \frac{dy}{dx}={\left(\sin x\right)}^x\left(x\cot x+1\ln \sin x\right)\end{array}$

At $x=0$ ,

  ${\sin }^0\left(0\right)\left(\left(\ln \left(\sin 0\right)\right)+0\cot 0\right)=\text{undefined}$

At $x=\frac{\pi }{2}$ ,

  ${\sin }^{\frac{\pi }{2}}\left(\frac{\pi }{2}\right)\left(\left(\ln \left(\sin \frac{\pi }{2}\right)\right)+\frac{\pi }{2}\cot \frac{\pi }{2}\right)=0$

Thus we can find $\frac{dy}{dx}$ of the given function.