1. ## Derivative help

differentiate e^x^ln(x)

u = x^ln(x)
derivative e^u= e^u

Now, how do i take derivative of x^ln(x) where a function is the power of another function?

2. Originally Posted by justin016
sorry wrong title, i meant derivative

differentiate e^x^ln(x)

u = x^ln(x)
derivative e^u= e^u

Lets assume you mean: differentiate $\displaystyle f(x)=e^{x^{\ln(x)}}$ with respect to $\displaystyle x$.

OK so you put $\displaystyle u=x^{\ln(x)}$, then:

$\displaystyle \frac{df}{dx}=\left(\frac{d}{du}e^u\right) \frac{du}{dx}=e^u \frac{d}{dx}(x^{\ln(x)})$

So now we see what you are asking, so try

$\displaystyle x^{\ln(x)}=(e^{\ln(x)})^{\ln(x)}=e^{(\ln(x))^2}$

CB

3. Originally Posted by justin016
sorry wrong title, i meant derivative

differentiate e^x^ln(x)

u = x^ln(x)
derivative e^u= e^u

Now, how do i take derivative of x^ln(x) where a function is the power of another function?
$\displaystyle \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}$

if $\displaystyle u = x^{\ln{x}}$

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

$\displaystyle \ln{u} = (\ln{x})^2$

$\displaystyle \frac{1}{u} \cdot \frac{du}{dx} = 2\ln{x} \cdot \frac{1}{x}$

$\displaystyle \frac{du}{dx} = \frac{2u \cdot \ln{x}}{x} = \frac{2x^{\ln{x}} \cdot \ln{x}}{x}$

so ...

$\displaystyle \frac{d}{dx} e^{x^{\ln{x}}} = e^{x^{\ln{x}}} \cdot \frac{2x^{\ln{x}} \cdot \ln{x}}{x}$