Derivative of $(a\,x)^{b\,x}$ is there any rule to differentiate the function $(a\,x)^{b\,x}$? 
I've got to find the derivative of $(x^2+1)^{\arctan x}$ and Wolfram|Alpha says the answer is 
$$\tan^{-1}(x) (x^2+1)^{\tan^{-1}(x)-1} \left(\frac{d}{dx}(x^2+1)\right)+\log(x^2+1) (x^2+1)^{\tan^{-1}(x)} \left(\frac{d}{dx}(\tan^{-1}(x))\right)$$
Is there any general rule to do that? 
Thanks.
 A: Assuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\ln(ax)bx} = e^{u(x)}$ and go from there).
A: Your derivative of $(x^2+1)^{\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\arctan x$ of  
$$\frac{d}{dx}\left(\left[ u(x)\right] ^{v(x)}\right)=v(x)\left[ u(x)\right]
^{v(x)-1}u^{\prime }(x)+\left( \ln u(x)\right) \left[ u(x)\right]
^{v(x)}v^{\prime }(x),$$
or omitting de variable $x$:
$$\left( u^{v}\right)^{\prime }=vu^{v-1}u^{\prime }+\left( \ln u\right) u^{v}v^{\prime }.$$
This can be derived observing that, since $u=e^{\ln u}$, we have $u^v=e^{v\ln u}$: 
$$\begin{eqnarray*}
\frac{d}{dx}\left( u^{v}\right)  &=&\frac{d}{dx}\left( e^{v\ln u}\right)  \\
&=&e^{v\ln u}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right)  \\
&=&u^{v}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right)  \\
&=&u^{v}\ln u\frac{dv}{dx}+u^{v}\frac{v}{u}\frac{du}{dx} \\
&=&u^{v}(\ln u)v'+u^{v-1}vu'.
\end{eqnarray*}$$
Another possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function 
$$z=f(u(x),v(x))=\left[ u(x)\right] ^{v(x)}$$
and determine its total derivative with respect to $x$:
$$\begin{eqnarray*}
\frac{dz}{dx} &=&\frac{d}{dx}\left( \left[ u(x)\right] ^{v(x)}\right) \\
&=&\frac{%
\partial z}{\partial u}\frac{du}{dx}+\frac{\partial z}{\partial v}\frac{dv}{%
dx} \\
&=&v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left[ u(x)\right]
^{v(x)}\left( \ln u(x)\right) v^{\prime }(x)
\end{eqnarray*}$$
because
$$\frac{\partial z}{\partial u}=\frac{\partial }{\partial u}\left(
u^{v}\right) =vu^{v-1}$$
and
$$\frac{\partial z}{\partial v}=\frac{\partial }{\partial v}\left(
u^{v}\right) =\frac{\partial }{\partial v}\left( e^{\ln u\cdot v}\right)
=e^{\ln u\cdot v}\ln u=u^{v}\ln u.$$
For $u(x)=ax,v(x)=bx$, we get
$$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right)
^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$$
A: HINT $\rm\ \ (F^G)'\ =\ (e^{G\:ln\ F})'\: =\ F^G\ (GF'/F + G'\ ln\ F)$
A: Not answering the math part, since Stijn has already done that, but if you click on the "show steps" button, Wolfram|Alpha shows you one possible path for the derivation.  I've included the image for the derivation of 
$\frac{\partial}{\partial x} ((a x)^{b x})$

A similar set of steps is supplied for the other derivative that you want to take: http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))
