Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}

share|improve this question
add comment

3 Answers

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 $h\to 0\implies 2h\ne0$

$=1$

share|improve this answer
add comment

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

share|improve this answer
    
... so what is $f'(0)$? –  akkkk Dec 16 '12 at 11:14
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.