Q:

Taking the derivative of $(1+x^2)^{(\sqrt{x})}$

As stated above, I'm having trouble taking the derivative of $(1+x^2)^{(\sqrt{x})}$. I know that I should somehow be using the exponential derivative form of $\dfrac{d}{dx} ( a^x ) = a^x\ln(a)$, but I can't quite figure how the product rule comes into play. 
Any help would be much appreciated!!

A:

We write
$$y = {\left( {1 + {x^2}} \right)^{\sqrt x }} = {e^{\sqrt x \ln \left( {1 + {x^2}} \right)}}.$$
So
$${y^/} = {e^{\sqrt x \ln \left( {1 + {x^2}} \right)}}{\left[ {\sqrt x \ln \left( {1 + {x^2}} \right)} \right]^/} = {e^{\sqrt x \ln \left( {1 + {x^2}} \right)}}\left[ {\frac{1}{{2\sqrt x }}\ln \left( {1 + {x^2}} \right) + \frac{{2x\sqrt x }}
{{1 + {x^2}}}} \right].$$

A:

$\begin{align}y=\left(1+x^2\right)^{\sqrt{x}}& \implies\ln y=\sqrt{x}\ln \left(1+x^2\right)\\&\implies \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{2x\sqrt{x}}{1+x^2}+\dfrac{\ln\left(1+x^2\right)}{2\sqrt{x}}\\&\implies\dfrac{dy}{dx}=\left(1+x^2\right)^{\sqrt{x}}\left(\dfrac{2x\sqrt{x}}{1+x^2}+\dfrac{\ln\left(1+x^2\right)}{2\sqrt{x}}\right)\end{align}$

