Derivative function went wrong I am trying to take the derivative of this function but I am facing some difficulties.
$$f(x)= e^{\ln(e^{7x^2+11})}$$
My answer was : $7e^{(7(x^2))}*14x$
 I cancelled the $\ln$ with the $e$ first, then I downgrade the $7$ and keep the $\exp$. as it is, after that I took the derivative of the $7x^2$ and the result was the one on top.
 A: If you are looking for the derivative of the following function
$$f(x)= e^{ln(e^{7x^2+11})}$$
then your first step would be correct. The $e$ and the $\ln$ "cancel" each other out and you are left with 
$$e^{7x^2+11}$$
Now, this is a classic case for the chain rule which states 
$$\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$$
Suppose our function is $z = e^{7x^2+11}$ and we want to find the derivative with respect to $x$ but our function is a composite function so we let the composite part equal $y$ which would be $y = 7x^2 + 11$. Now, we use the chain rule so we have
\begin{align}
\frac{dz}{dx} &= \frac{dz}{dy} \cdot \frac{dy}{dx} \\
\frac{dz}{dx} &= \frac{d}{dy} e^y \cdot \frac{d}{dx}7x^2+11 \\
&= e^y \cdot 14x \\
&= e^{7x^2+11} \cdot 14x
\end{align}
So, our answer is $e^{7x^2+11} \cdot 14x$. I feel that you are making an error when you talk about downgrading the $7$, I don't quite understand why you would need to do that. If you have any further questions then please don't hesitate to ask. 
A: $$f(x)= e^{\ln(e^{7x^2+11})}=\exp(\ln(\exp(7x^2+11)))$$
So
$$f(x)= \exp(7x^2+11), \implies f'(x)=14x\exp(7x^2+11)$$
