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Question

Use logarithmic differentiation to find the derivative of the function. y=xlnxy=x^{\ln x}

Solution

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First, we will take the natural logarithm on both sides of the given equation and get

lny=lnxlnx\ln y=\ln x^{\ln x}

Using the fact that lnab=blna\pmb{\ln a^b=b\ln a}, we have

lny=lnxlnx\ln y=\ln x\cdot \ln x

If we differentiate both sides of this equation, we get

(lny)=(lnxlnx)yy=(lnx)lnx+lnx(lnx)yy=lnxx+lnxxy=y[lnxx+lnxx]y=2lnxxxlnxy(x)=2lnxxlnx1\begin{align*} (\ln y)^\prime&=(\ln x\cdot \ln x)^\prime\\ \frac{y^\prime}{y}&=(\ln x)^\prime\cdot \ln x+\ln x \cdot (\ln x)^\prime\\ \frac{y^\prime}{y}&=\frac{\ln x}{x}+\frac{\ln x}{x}\\ y^\prime&=y\cdot \left[\frac{\ln x}{x}+\frac{\ln x}{x}\right]\\ y^\prime&=\frac{2\ln x}{x}\cdot x^{\ln x}\Rightarrow\boxed{\pmb{y^\prime(x)=2\ln x\cdot x^{\ln x-1}}} \end{align*}

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