Homework Help: 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.