Finding the derivative of $(3x^2 +5)^{\arctan x}$ 
Can someone please explain how the power rule, brings natural log into the equation? Why is the derivative not $\arctan(x)(3x^2+5)^{\arctan(x)-1}$?
 A: You can write $a^b$ as $e^{b \ln a}$. So in this case we have $$(3x^2 + 5)^{\arctan x} = \exp(\arctan x \ln (3x^2+5))$$ and we know that the derivative of $\exp f(x)$ is $f'(x) \exp f(x)$ by the chain rule. 
As for why the derivative of $f^g$ is not simply $gf^{g-1}$ for differentiable functions $f$ and $g$ in general, well - it's simply not true. We can provide many counterexamples. The 'power rule' only works if want to differentiate something of the form $x^c$ where $c$ is emphatically a constant. 
A: Recall that $e^{\log(x)}=x$.  Hence, $$f(x)^{g(x)}=e^{\log(f(x)^{g(x)})}=e^{g(x)\log(f(x))}\tag 1$$
Using the chain rule in $(1)$, we can assert that 
$$\frac{df(x)^{g(x)}}{dx}=e^{g(x)\log(f(x)}\frac{d(g(x)\log(f(x))}{dx}\tag 2$$
Then, let $f(x)=3x^2+5$ and $g(x)=\arctan(x)$ in $(2)$ and proceed.
A: For this kind of monsters, logarithmic differentiation is your best friend.
Making the problem more general, consider $$y={f(x)}^{g(x)}\implies \log(y)=g(x) \log(f(x))$$ Differentiate both sides $$\frac {y'}y=g(x)\frac{f'(x)}{f(x)}+g'(x)\log\left({f(x)} \right)$$ Now, use $$y'=y\times \frac {y'}y$$
