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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}

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up vote 1 down vote accepted

$$\lim_{h\to 0}\frac{e^{2(x+h)}-e^{2x}}h=2e^{2x}\lim_{h\to 0}\left(\frac{e^{2h}-1}{2h}\right)=2e^{2x}$$ 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 $h\to 0\implies 2h\ne0$


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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, $$f^{\prime}(x)=e^{2x}\lim_{h\to 0}\frac{e^{2h}-1}{h}=2e^{2x}\lim_{u\to 0}\frac{e^{u}-1}{u}=2e^{2x}g^{\prime}(0)=2e^{2x}$$

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... so what is $f'(0)$? – akkkk Dec 16 '12 at 11:14

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: $$\lim_{h\to0}e^{2x}\frac{e^{2h} - 1}{h}= \lim_{u\to0}2e^{2x}[\frac{u}{\ln(1+u)}]$$ $$= 2e^{2x}\lim_{u\to0}[\frac{1}{\ln(1+u)^\frac{1}{u}}] =$$ $$= 2e^{2x}[\frac{1}{\ln\lim_{u\to0}(1+u)^\frac{1}{u}}] =$$ $$= 2e^{2x}[\frac{1}{\ln e}] = 2e^{2x}\cdot 1 =$$ $$= 2e^{2x}$$

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