Question

Use logarithmic differentiation to find the derivative of thefunction.

49. \(y = {x^x}\)

Short answer

The answer is \(\frac{{dy}}{{dx}} = {x^x}\left( {1 + \ln x} \right)\).

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

Taking log both side to get,

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

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

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

Hence \(\frac{{dy}}{{dx}} = {x^x}\left( {1 + \ln x} \right)\).