# Derivative of 2^x

1. Jan 5, 2010

### CrimeBeats

Hi, i've been trying to find the derivative of 2^x and i got stuck here:
Lim (2^x - 1)/x
x=>0

How can i solve this limit (with steps please)
ps. im only 17

2. Jan 5, 2010

### LCKurtz

Have you had the same limit for ex? What you want to do is notice that

2x = (eln(2))x = exln(2)

$$\lim_{x\rightarrow 0}\frac{e^{x\ln 2} - 1}{x}$$

Now let u = x ln(2) and use what you know about ex.

 Corrected typo

Last edited: Jan 5, 2010
3. Jan 5, 2010

### yungman

I have been doing it this way all the time which I thought it is very straight forward:

$$y=2^{x}\Rightarrow ln(y)=xln(2)$$

$$\Rightarrow \frac{1}{y}\frac{dy}{dx}=ln(2)$$

$$\Rightarrow dy=d[2^{x}]=yln2dx=2^x ln(2)dx$$

Last edited: Jan 5, 2010
4. Jan 5, 2010

### HallsofIvy

No, but probably a typo. You have to multiply both sides by y dx, not x dx.
$dy= yln(2)dx= 2^x ln(2) dx$.

5. Jan 5, 2010

### Anonymous217

With $$e^{ln(2)x}$$, you can easily take the derivative of that. Just don't simplify the $$e^{ln2}$$ part. You should get $$ln(2)e^{ln(2)x}$$ as the numerator while the denominator becomes 1. Now do this expression with $$\lim_{x->0}$$. By the way, LCKurtz made a slight typo.

6. Jan 5, 2010

### yungman

My finger is get too far ahead of me!!! And I have destroy evidence already!!!

7. Jan 7, 2010

### CrimeBeats

I guess the answer is ln(2) * 2^x
No wonder i couldnt solve it before, they thaught us nothing about e

Depending on what you've said this means the derivative of e^x is e^x.. right?

8. Jan 7, 2010

### LCKurtz

Yes. Most calculus books do that first before trying other bases.

9. Jan 8, 2010

### HallsofIvy

This problem might have been given in preparation for introducing "e".

The derivative of $2^x$ is
$$\lim_{h\to 0}\frac{2^{x+y}- 2^x}{h}= \lim_{h\to 0}\frac{2^x2^h- 2^x}{h}$$
$$= \lim{h\to 0}\left(\frac{2^h- 1}{h}\right)2^x[/itex] [tex]= \left(\lim_{h\to 0}\frac{2^h- 1}{h}\right)2^x$$
That is, of course, a constant times $2^x$.

Similarly, the derivative of $3^x$ is
$$= \left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)3^x$$
a constant times $3^x$

In that same way you can show that the derivative of $a^x$, for a any positive real number, is $C_a a^x$.
Further, by numerical approximations, you can show that $C_2$ is less than 1 and $C_3$ is greater than 1. There exists, then, a number, a, between 2 and 3 such that $C_a= 1$. If we call that number "e", then the derivative of $e^x$ is just $e^x$ itself.