Find Derivative of y=2^x Using Definition

[hackensack]

I need to find the derivitive of y=2^x using the definition of derivitive.

 

[Hurkyl]

What have you done so far? What does the definition of derivative say?

 

[HallsofIvy]

This was also posted in the calculus section and there are about 10 replies there.

 

[mathwonk]

I doubt this is a suitable problem for a novice. even showing convergence is tough. i will look at the other posted answers. there is a good reason people start from the integral definition of ln(x) to derive this result.

 

[HallsofIvy]

If f(x)= a^x, the f(x+h)= a^a+x= a^xa^h so
f(a+ h)- f(a)= a^x(a^h- 1).

The derivative is lim (f(x+h)- f(x))/h= a^xlim {(a^h-1)/h}. Notice that that is a^x time a limit that is independent of x. That is, as long as the derivative exists, it is a^x times a constant. The problem is showing that the lim{(a^h-1)/h} EXISTS! And then showing that, if a= 2, that limit is ln(2).

Showing that that limit exists is sufficiently non-trivial that many people (myself included), as mathwonk said, prefer to define ln(x) as the integral, from 1 to x of (1/t)dt. From that, it is possible to prove all properties of ln(x) including (trivially) that the derivative is 1/x. Defining e^x as the inverse function of ln(x) leads to all the properties of e^x (including the fact that it is some number to a power!), in particular that its derivative is e^x itself and, from that, that the derivative of a^x is (ln a) a^x.

 

[quantitative]

dunno if I'm missing the point here but...

write
y=2^x
as
y=exp(x.ln2)
=>
y'=ln2.exp(x.ln2)

 

[cepheid]

No, you did exactly what HallsofIvy was advocating, he was just pointing out that the question asked for it to be solved using the definition of a derivative, which makes things much harder. Easier to approach things from the other way, starting by defining the integral of 1/x.

 

[brout]

"The derivative is lim (f(x+h)- f(x))/h= axlim {(ah-1)/h}. Notice that that is ax time a limit that is independent of x. That is, as long as the derivative exists, it is ax times a constant. The problem is showing that the lim{(ah-1)/h} EXISTS! And then showing that, if a= 2, that limit is ln(2)."

tell me if I'm wrong, but it doesn't seems so hard to determine this limit..
(a^h-1)/h = (exp (h*ln(a) )-1) / h
= ( 1 + h*ln(a) + o(h*ln(a)) - 1 ) / h h->0
= ln(a) + o(ln(a))
so lim (a^h-1)/h = ln(a) ...

 

[HallsofIvy]

"The derivative is lim (f(x+h)- f(x))/h= axlim {(ah-1)/h}. Notice that that is ax time a limit that is independent of x. That is, as long as the derivative exists, it is ax times a constant. The problem is showing that the lim{(ah-1)/h} EXISTS! And then showing that, if a= 2, that limit is ln(2)."

tell me if I'm wrong, but it doesn't seems so hard to determine this limit..
(a^h-1)/h = (exp (h*ln(a) )-1) / h
= ( 1 + h*ln(a) + o(h*ln(a)) - 1 ) / h h->0
= ln(a) + o(ln(a))
so lim (a^h-1)/h = ln(a) ...

Yes, assuming that you know "(exp (h*ln(a) )-1) / h= ( 1 + h*ln(a) + o(h*ln(a)) - 1 ) / h " its easy to do it. Proving what you assumed is the hard part!