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

Jul 4, 2018

$f ' \left(x\right) = 2 {x}^{\ln \left(x\right)} \cdot \left(\ln \frac{x}{x}\right)$

#### Explanation:

Taking the logarithm on both sides we get

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

$f ' \left(x\right) = 2 {x}^{\ln \left(x\right)} \cdot \left(\ln \frac{x}{x}\right)$

$\setminus \frac{d}{\mathrm{dx}} \left({x}^{\setminus \ln x}\right) = {x}^{\setminus \ln x} \left(\setminus \frac{1}{{x}^{2}} + \setminus \ln x\right)$

#### Explanation:

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

$\setminus \frac{d}{\mathrm{dx}} \left({x}^{\setminus \ln x}\right)$

$= {x}^{\setminus \ln x - 1} \setminus \frac{d}{\mathrm{dx}} \left(\setminus \ln x\right) + {x}^{\setminus \ln x} \setminus \ln x \setminus \frac{d}{\mathrm{dx}} \left(x\right)$

$= {x}^{\setminus \ln x - 1} \left(\setminus \frac{1}{x}\right) + {x}^{\setminus \ln x} \setminus \ln x$

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

$= {x}^{\setminus \ln x} \left(\setminus \frac{1}{{x}^{2}} + \setminus \ln x\right)$