Derivative of 2^x: Solving Limits at 17

[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. I am only 17

 

[LCKurtz]

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. I am only 17

Have you had the same limit for e^x? What you want to do is notice that

2^x = (e^ln(2))^x = e^xln(2)

Now your limit becomes:

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

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

[Edit] Corrected typo

 

[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

 

[HallsofIvy]

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}]=xln(2)dx

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.

 

[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.

 

[yungman]

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.

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

 

[CrimeBeats]

I guess the answer is ln(2) * 2^x
No wonder i couldn't 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?

 

[LCKurtz]

I guess the answer is ln(2) * 2^x
No wonder i couldn't 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?

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

 

[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]<br /> = \left(\lim_{h\to 0}\frac{2^h- 1}{h}\right)2^x<br /> That is, of course, a constant times 2^x.<br /> <br /> Similarly, the derivative of 3^x is<br /> = \left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)3^x<br /> a constant times 3^x<br /> <br /> In that same way you can show that the derivative of a^x, for a any positive real number, is C_a a^x.<br /> 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 &quot;e&quot;, then the derivative of e^x is just e^x itself.