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2^x derivitive

by hackensack
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hackensack
#1
May19-04, 04:10 PM
P: 2
I need to find the derivitive of y=2^x using the definition of derivitive.
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Hurkyl
#2
May19-04, 04:30 PM
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What have you done so far? What does the definition of derivative say?
HallsofIvy
#3
May21-04, 05:28 AM
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This was also posted in the calculus section and there are about 10 replies there.

mathwonk
#4
Sep21-04, 08:27 PM
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2^x derivitive

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
#5
Sep21-04, 09:17 PM
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If f(x)= ax, the f(x+h)= aa+x= axah so
f(a+ h)- f(a)= ax(ah- 1).

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).

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 ex as the inverse function of ln(x) leads to all the properties of ex (including the fact that it is some number to a power!), in particular that its derivative is ex itself and, from that, that the derivative of ax is (ln a) ax.
quantitative
#6
Oct15-04, 07:23 AM
P: 3
dunno if i'm missing the point here but...

write
y=2^x
as
y=exp(x.ln2)
=>
y'=ln2.exp(x.ln2)
cepheid
#7
Oct16-04, 02:53 AM
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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
#8
Oct29-04, 12:11 PM
P: 5
"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
#9
Oct29-04, 01:32 PM
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Quote Quote by 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) .......
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!


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