Calculating the definition of a derivative: 2^x

[Potatochip911]

Homework Statement

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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.

Homework Equations

y=2^x

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.

 

[vela]

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?
$$\lim_{h\to 0}\frac{e^h - 1}h = 1$$

 

[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.

 

[vela]

Honestly I'm not entirely sure what I'm allowed to use.

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

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?

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 know how to derive e^x using the chain rule but I'm not sure how I could incorporate that into my limit.

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

 

[HallsofIvy]

"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).)