Question

Use logarithmic differentiation to find the derivative of thefunction.

51. \(y = {x^{sinx}}\)

Short answer

The answer is \(\frac{{dy}}{{dx}} = {x^{\sin x}}\left( {\frac{{\sin x}}{x} + \ln x\cos x} \right)\).

Step-by-step solution

Write the formula of the logarithmic differentiation and derivative of logarithmic functions

Derivatives of logarithmic functions:\(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)

Logarithmic differentiation:If \(y = {x^n}\) then\(\ln y = n\ln x\)

Finding the derivative of the function

Given that \(y = {x^{\sin x}}\).

Taking log both side to get,

\(\ln y = \sin x\ln x\)

Now differentiating with respect to \(x\) we get,

\(\begin{aligned}{c}\frac{1}{y}\frac{{dy}}{{dx}}&= \frac{d}{{dx}}\left( {\sin x\ln x} \right)\\\frac{{dy}}{{dx}}&= y\left( {\sin x\frac{d}{{dx}}\ln x + \ln x\frac{d}{{dx}}\sin x} \right)\\&= {x^{\sin x}}\left( {\frac{{\sin x}}{x} + \ln x\cos x} \right)\end{aligned}\)

Hence \(\frac{{dy}}{{dx}} = {x^{\sin x}}\left( {\frac{{\sin x}}{x} + \ln x\cos x} \right)\).