# The derivative of 2^x

1. May 24, 2012

### robertjford80

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

I'm trying to find the derivative of 2^x by hand so that I can better understand the number e. This video

says the answer is .69. I can't figure out how to get that

2. Relevant equations

lim h -> 0 [f(x0+h) - f(x0)]/h

3. The attempt at a solution

So let's take point (3,8)

[2(3+h)^3 - 4(3)^3]/h

= [54 + 2h^3 - 108]/h

= 54

not exactly .69

2. May 24, 2012

### Jorriss

I am not sure what you are doing here - but that looks incorrect.

The easiest way to do this derivative is to write y = 2^x and take the natural log of both sides and then differentiate.

3. May 24, 2012

### robertjford80

If I do that, I get

ln y = x ln 2

Two unknowns and one equation, can't be solved.

4. May 24, 2012

### Jorriss

No... you take the derivative with respect to x of both sides.

5. May 24, 2012

### robertjford80

Then I will still have an x in my solution, not .69

6. May 24, 2012

### robertjford80

wait, the natural log of 2 is .69, ok, I get it now.

7. May 24, 2012

### Jorriss

Are you sure? Because the derivative of 2^x is not .69. It's (ln2)*2^x. It is only .69 at x=0.

8. May 24, 2012

### robertjford80

well, i get enough to satisfy me for the moment.

9. May 24, 2012

### Jorriss

well, ok then.

10. May 24, 2012

### clamtrox

This is wrong on so many levels... You should really read through the more basic stuff again...

So if f(x) = 2x, then f(x0+h)-f(x0) = 2(x0+h)-2x0

Next, you don't pick a random value for h... Instead, you take the limit as h tends to zero of (f(x+h)-f(x))/h

11. May 24, 2012

### sharks

Or you could just learn the general formula: $\frac{d(a^x)}{dx}=a^x.\ln a$ where a is any real constant.

12. May 24, 2012

### HallsofIvy

Staff Emeritus
Your whole question doesn't make much sense. The derivative of $2^x$ is a function of x, not a number. Since, in fact, the derivative of $2^x$ is the function $ln(2)2^x$, and ln(2) is (approximately) 0.69, the derivative of $2^x$ at x= 0 is (approximately) 0.69.

Pretty much everything here is wrong. For one thing, $(x+ h)^3= x^3+ 3x^2h+ 3xh^3+ h^3$, NOT "$x^3+ h^3$". But where did that $2(3+h)^3$ come from anyway? The difference quotient for $2^x$ would be
$$\frac{2^{x+ h}- 2^x}{h}$$
not what you have. (Which looks like it would be for $2x^3$ if that "4" were a "2".)

We could then write $2^{x+y}= 2^x2^h$ and factor $2^x$ out. The difference quotient becomes
$$2^x \frac{2^h- 1}{h}$$
so that the derivative, the limit, as h goes to 0, of that, is $2^x$ times the limit of that last fraction. In fact, it is easy to see that the derivative of $a^x$ is just [itex]a^x[itex] itself times the limit of
$$\frac{a^h- 1}{h}$$

"e" happens to have that limit equal to 1. And one can use the properties of [itex]e^x[itex] to show that
$$\lim_{h\to 0}\frac{a^h- 1}{h}= ln(a)$$