How to find $f'$ from the definition of derivative? How to find the derivative of this function: $f(x) = e^{2x}$ - using definition of derivative:
\begin{equation}
    f'(x) = \lim_{h\to0}\dfrac{f(x + h) - f(x)}{h}
\end{equation}
 A: $$\lim_{h\to 0}\frac{e^{2(x+h)}-e^{2x}}h=2e^{2x}\lim_{h\to 0}\left(\frac{e^{2h}-1}{2h}\right)=2e^{2x}$$ as 
$\lim_{h\to 0}\left(\frac{e^{2h}-1}{2h}\right)$
$=\lim_{h\to 0}\left(\frac{1+\frac{2h}{1!}+\frac{(2h)^2}{2!}+\cdots-1}{2h}\right)$
$=\lim_{h\to 0}\left(1+\frac{2h}{2!}+\frac{(2h)^2}{3!}+\cdots\right)$ as $h\to 0\implies 2h\ne0$
$=1$
A: By definition
$$f^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{e^{2(x+h)}-e^{2x}}h=\lim_{h\to 0}\frac{e^{2x}e^{2h}-e^{2x}}h=e^{2x}\lim_{h\to 0}\frac{e^{2h}-1}{h}$$
We must compute
$$\lim_{h\to 0}\frac{e^{2h}-1}{h}$$
With the change of variables $u=2h$, this becomes
$$2\lim_{u\to 0}\frac{e^{u}-1}{u}$$
which is nothing but $g^{\prime}(0)$ where $g(x)=e^x$. Therefore,
$$f^{\prime}(x)=e^{2x}\lim_{h\to 0}\frac{e^{2h}-1}{h}=2e^{2x}\lim_{u\to 0}\frac{e^{u}-1}{u}=2e^{2x}g^{\prime}(0)=2e^{2x}$$
A: Using the definition of derivative:
$$\lim_{h\to0}\frac{f(x + h) - f(x)}{h} = \lim_{h\to0}\frac{e^{2x+2h} - e^{2x}}{h}=$$
$$= \lim_{h\to0}e^{2x}\frac{e^{2h} - 1}{h}$$
Let's define $u$ so that $u=e^{2h}-1$,when $h\to 0$, $u \to 0$ too.
So if we continue, we get:
$$\lim_{h\to0}e^{2x}\frac{e^{2h} - 1}{h}= \lim_{u\to0}2e^{2x}[\frac{u}{\ln(1+u)}]$$
$$= 2e^{2x}\lim_{u\to0}[\frac{1}{\ln(1+u)^\frac{1}{u}}] =$$
$$= 2e^{2x}[\frac{1}{\ln\lim_{u\to0}(1+u)^\frac{1}{u}}] =$$
$$= 2e^{2x}[\frac{1}{\ln e}] = 2e^{2x}\cdot 1 =$$
$$= 2e^{2x}$$
