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
$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right) ^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$