Question

Use logarithmic differentiation to find the derivative of thefunction.

50. \(y = {x^{\frac{1}{x}}}\)

Short answer

The answer is \(\frac{{dy}}{{dx}} = {x^{\frac{1}{x}}}\left( {\frac{{1 - \ln x}}{{{x^2}}}} \right)\).

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 \(y = {x^{\frac{1}{x}}}\).

Taking log both side to get,

\(\ln y = \frac{1}{x}\ln x\)

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

\(\begin{aligned}{c}\frac{1}{y}\frac{{dy}}{{dx}}&= \frac{d}{{dx}}\frac{1}{x}\ln x\\\frac{{dy}}{{dx}}&= {x^{\frac{1}{x}}}\left( {\frac{1}{x}\frac{d}{{dx}}\ln x + \frac{d}{{dx}}\frac{1}{x}\ln x} \right)\\&= {x^{\frac{1}{x}}}\left( {\frac{1}{{{x^2}}} - \frac{1}{{{x^2}}}\ln x} \right)\\&= {x^{\frac{1}{x}}}\left( {\frac{{1 - \ln x}}{{{x^2}}}} \right)\end{aligned}\)

Hence \(\frac{{dy}}{{dx}} = {x^{\frac{1}{x}}}\left( {\frac{{1 - \ln x}}{{{x^2}}}} \right)\).