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?
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
$\displaystyle \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}$
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 ...
$\displaystyle \frac{d}{dx} e^{x^{\ln{x}}} = e^{x^{\ln{x}}} \cdot \frac{2x^{\ln{x}} \cdot \ln{x}}{x}$