# How do you differentiate f(x)=(ln(x))^x ?

Dec 12, 2015

Take the natural log of both sides, then use implicit differentiation ...

#### Explanation:

Take natural log of both sides, then use the property of logs :

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

Now, using implicit differentiation, product and chain rules ...

$\left(\frac{1}{y}\right) y ' = \ln \left(\ln x\right) + \left(\frac{x}{\ln} x\right) \times \frac{1}{x} = \ln \left(\ln x\right) + \frac{1}{\ln} x$

Finally, solve for $y '$

$y ' = y \left[\ln \left(\ln x\right) + \frac{1}{\ln} x\right] = {\left(\ln x\right)}^{x} \left[\ln \left(\ln x\right) + \frac{1}{\ln} x\right]$

hope that helped

Dec 12, 2015

(IGNORE)

#### Explanation:

Rewrite $f \left(x\right)$ using the properties of logarithms.

$f \left(x\right) = x \ln \left(x\right)$

Now, to find $f ' \left(x\right)$, use the product rule.

$f ' \left(x\right) = \ln \left(x\right) \frac{d}{\mathrm{dx}} \left[x\right] + x \frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right]$

Find each derivative.

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right] = \frac{1}{x}$

Plug the derivatives back in.

$f ' \left(x\right) = 1 \left(\ln \left(x\right)\right) + x \left(\frac{1}{x}\right)$

$f ' \left(x\right) = \ln \left(x\right) + 1$