How do you find the derivative of $x^ln x$ ?

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Answer 1 · 2018-07-04T18:55:45

$f'(x)=2x^(ln(x))*(ln(x)/x)$

Taking the logarithm on both sides we get

$ln(f(x))=ln(x)*ln(x)$
differentiating with respect to $x$ :
$1/f(x)*f'(x)=ln(x)/x+ln(x)/x)$
so we get

$f'(x)=2x^(ln(x))*(ln(x)/x)$

Answer 2 · 2018-07-04T18:59:17

$\frac{d}{dx}(x^{\ln x})=x^{\lnx}(\frac{1}{x^2}+\ln x)$

For a positive real number $x$ , applying chain rule of differentiation as follows

$\frac{d}{dx}(x^{\ln x})$

$=x^{\lnx-1}\frac{d}{dx}(\ln x)+x^{\lnx}\ln x\frac{d}{dx}(x)$

$=x^{\lnx-1}(\frac{1}{x})+x^{\lnx}\ln x$

$=x^{\lnx}\frac{1}{x^2}+x^{\lnx}\ln x$

$=x^{\lnx}(\frac{1}{x^2}+\ln x)$

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