Question

Use logarithmic differentiation to find the derivative of the function.

54.\(y = {\left( {sinx} \right)^{lnx}}\)

Short answer

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

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 \(y = {\left( {\sin x} \right)^{\ln x}}\).

Taking log both side to get,

\(\ln y = \ln x\ln \left( {\sin x} \right)\)

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

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

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