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: $\begin{align}y=\left(1+x^2\right)^{\sqrt{x}}& \implies\ln y=\sqrt{x}\ln \left(1+x^2\right)\\&\implies \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{2x\sqrt{x}}{1+x^2}+\dfrac{\ln\left(1+x^2\right)}{2\sqrt{x}}\\&\implies\dfrac{dy}{dx}=\left(1+x^2\right)^{\sqrt{x}}\left(\dfrac{2x\sqrt{x}}{1+x^2}+\dfrac{\ln\left(1+x^2\right)}{2\sqrt{x}}\right)\end{align}$
A: Hint:
$$(1+x^2)^{\sqrt x} = e^{\ln\left((1+x^2)^{\sqrt x}\right)} = e^{\sqrt x\ln(1+x^2)}$$
Differentiating that should be easier, since for a differentiable function $f$, you have $$\left(e^{f(x)}\right)' = f'e^{f(x)}$$
(using the chain rule)
A: Hint: $\quad\bigg[\Big(1+x^2\Big)^{\sqrt x}~\bigg]'\quad=\quad\Big[a^{\sqrt x}\Big]'_{\large a=1+x^2}\quad+\quad\bigg[\Big(1+x^2\Big)^n\bigg]'_{\large n=\sqrt x}\qquad$ where 
$\Big[a^{\sqrt x}\Big]'~=~a^{\sqrt x}\cdot\ln a\cdot\Big(\sqrt x\Big)'\qquad$ and $\qquad\bigg[\Big(1+x^2\Big)^n\bigg]'~=~n~\Big(1+x^2\Big)^{n-1}\cdot\Big(1+x^2\Big)'$.
