Find derivative using the limit definition for the function $f(x) = e^{2(x+1)}$ Use the limit definition of derivative to find $\frac{\mathrm d}{\mathrm dx}f(x)$ for the function $f(x)= e^{2(x+1)}$.
I know it's going to be $2e^{2x+2}$. I can solve it normally, I just don't know how to solve it using the definition, please help. 
 A: Following the definition we get :
\begin{align*}
f'(x) ={}& \lim_{h\rightarrow 0} {f(x+h)-f(x)\over h} \\
{}={}& \lim_{h\rightarrow 0} {e^{2(x+h+1)}-e^{2(x+1)}\over h} \\
{}={}& \lim_{h\rightarrow 0} {e^{2h}e^{2(x+1)}-e^{2(x+1)}\over h} \\
{}={}& e^{2(x+1)} \lim_{h\rightarrow 0} {e^{2h}-1\over h} \\
{}={}& e^{2(x+1)} \lim_{h\rightarrow 0} {1 \over h} \left(2h+{(2h)^2 \over 2!} +{(2h)^3 \over 3!} + \cdots \right) \\
{}={}& 2 \cdot e^{2(x+1)} \\
\end{align*}
A: $$\frac{\mathrm d}{\mathrm dx}\left[e^{2(x+1)}\right] $$
$$= \lim\limits_{h\to 0} \frac{e^{2(x+h+1)}-e^{2(x+1)}}{h}$$
$$= \lim\limits_{h\to 0} \frac{\left(e^{2}\right)^{(x+h+1)}-\left(e^{2}\right)^{(x+1)}}{h}$$
$$= \lim\limits_{h\to 0} \frac{\left(e^{2}\right)^{(x+1)}\left(e^{2}\right)^h-\left(e^{2}\right)^{(x+1)}}{h}$$
$$= \left(e^{2}\right)^{(x+1)}\lim\limits_{h\to 0} \frac{\left(e^{2}\right)^h-1}{h}$$
$$= 2e^{2(x+1)}\lim\limits_{h\to 0} \frac{e^{2h}-1}{2h}$$
Let $t=2h$, so now we have
$$2e^{2(x+1)}\lim\limits_{t\to 0} \frac{e^{t}-1}{t}$$
$$=2e^{2(x+1)}\ln e$$
$$=2e^{2(x+1)}$$
