# How do you differentiate y=x^lnx?

##### 1 Answer
Apr 18, 2016

$y ' = 2 {x}^{\ln \left(x\right) - 1} \ln \left(x\right)$

#### Explanation:

Using logarithmic and implicit differentiation, take the natural logarithm of both sides.

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

Simplify the right hand side using the rule: $\ln \left({a}^{b}\right) = b \cdot \ln \left(a\right)$

$\ln \left(y\right) = \ln \left(x\right) \cdot \ln \left(x\right)$

$\ln \left(y\right) = {\left(\ln \left(x\right)\right)}^{2}$

Take the derivative of both sides. Recall that the chain rule is in effect on both sides.

$\frac{y '}{y} = 2 \ln \left(x\right) {\overbrace{\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right)}}^{= 1 / x}$

$\frac{y '}{y} = \frac{2 \ln \left(x\right)}{x}$

Now, multiply both sides by $y = {x}^{\ln} \left(x\right)$ to solve for $y '$.

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

Note that we can simplify this be writing ${x}^{\ln} \frac{x}{x}$ as ${x}^{\ln} \frac{x}{x} ^ 1 = {x}^{\ln \left(x\right) - 1}$. Thus,

$y ' = 2 {x}^{\ln \left(x\right) - 1} \ln \left(x\right)$