Calculating the definition of a derivative: 2^x

1. Dec 5, 2014

Potatochip911

1. The problem statement, all variables and given/known data

I'm supposed to find the derivative of 2^x using the definition of a derivative. I am really confused as to how I can factor out the h.

2. Relevant equations
y=2^x

3. The attempt at a solution
limit as h->0 in all of these, I don't want to write it out because it's going to look even worse (didn't want to learn latex right this moment).
dy/dx=(2^(x+h)-2^x)/h
dy/dx=(2^x*(2^h-1))/h
This is as far as I got and I'm not even sure if it was useful to factor out 2^x
Hopefully there is a rule for exponents that I'm not aware of that can be used for this.

2. Dec 5, 2014

vela

Staff Emeritus
Use the fact that $2^h = e^{h \log 2}$ (where $\log$ is the natural log).

I guess you need to know some other stuff as well. Without knowing what you're allowed to use, it's hard to give you advice. Do you know the derivative of $e^x$ for example? Or the following limit?
$$\lim_{h\to 0}\frac{e^h - 1}h = 1$$

3. Dec 6, 2014

Potatochip911

Honestly I'm not entirely sure what I'm allowed to use, we went over the proof for the derivative of e but we were told that we wouldn't need to know it, this question is from a previous year so maybe that year the curriculum was different? I know how to derive e^x using the chain rule but I'm not sure how I could incorporate that into my limit.

4. Dec 6, 2014

vela

Staff Emeritus
Well, that's one thing you'll need to figure out.

You can solve this problem pretty easily if you can use the fact that the derivative of $e^x$ is $e^x$. You don't need to know how to prove it.

I think you mean you know how to differentiate $e^x$. Derive doesn't mean "find the derivative of."

5. Dec 6, 2014

HallsofIvy

Staff Emeritus
"Using the definition of a derivative" means using the difference quotient:
$$\frac{df}{dx}= \lim_{h\to 0} \frac{f(x+ h)- f(x)}{h}$$

$$\frac{2^{x+h}- 2^x}{h}= \frac{2^x(2^h)- 2^x}{h}= 2^x \frac{2^h- 1}{h}$$

So $\frac{d 2^x}{dx}= 2^x\left(\lim_{h\to 0} \frac{2^h- 1}{h}\right)$
That is obviously $2^x$ times a constant. How to find that constant, and if you can, depends on what other knowledge you have.
($2^x= e^{ln(2^x)}= e^{x ln(2)}$ so if you know how to differentiate $e^x$ you can show that constant is ln(2).)