Question

how would I take derivative of 2^(x)

Answer 1

set it ln it

Answer 2

2^x ln(2)

Answer 3

think of it like e^x

e^x' = e^x ln(e)

Answer 4

rewrite it as
\[e^{x\ln(2)}\] and take the derivative of that

Answer 5

Write 2^x as exp(ln(2^x)) = exp(ln2.x) , i.e, e^(ln2.x)

So the derivative is ln2.e^(ln2.x) = ln 2 . 2^x

Answer 6

although it is probably just best to remember that the derivative of
\[b^x\] is
\[\ln(b)\times b^x\]

Answer 7

why does e^x gets its own special ruling?

Answer 8

alternatively you can always write
\[\frac{2^{x+h}-2^x}{h}=\frac{2^x\times 2^h-2^x}{h}=\frac{2^x(2^h-1)}{h}=2^x\times \frac{2^h-1}{h}\] then take
\[2^x\times \lim_{h\rightarrow 0}\frac{2^h-1}{h}\] and some convince your self that
\[\lim_{h\rightarrow 0}\frac{2^h-1}{h}=\ln(2)\]

Answer 9

By definition of exp as a function.

The usual way this is done in a careful modern development of calculus is to say exp is the inverse function of ln where

\[\ln x = \int\limits_{1}^{x} dt/ t\]

if you define exp that way, then you can show that d/dt (exp(t)) = exp(t).

Answer 10

e gets special dispensation depending on your definition. one definition is that e is the number such that
\[\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1\]

Answer 11

[
Let f = ln, then inverse f, call it g, g = exp. Now g'(x) = 1/f'(g(x))

What is f'? f'(x) = 1/x. So, g'(x) = 1/(1/(g(x)) = g(x)
]

Answer 12

but doesnt that fit your other part as well?

\[\lim_{h\rightarrow 0}\frac{e^h-1}{h}=ln(e)=1\]

Answer 13

another is what jamesj says although it is not going to be convincing if you are starting calc and learning derivatives, because you certainly will not have seen integrals!

Answer 14

No, when you introduce this in HS you use a kind of intuitive logic and you're overly concerned about consistency.

But if you want to define these things carefully and make sure everything actually works consistently, then what I've laid out is a very good way to do it.

Answer 15

@amistre you can show easily enough that
\[\lim_{x\rightarrow 0}\frac{2^x-1}{x}<1\] and
\[\lim_{x\rightarrow 0}\frac{3^x-1}{x}>1\] and the say e is the number for which
\[\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1\]

Answer 16

that will mean that the derivative of exp is exp

Answer 17

The big problem with your proposal is you need to show that the function defined by

f(x) = lim_{t->0} (x^t - 1)/t

is (a) well defined, (b) continuous and (c) only has one zero between 2 and 3. That doesn't look so straight-forward to me.

Answer 18

that is true enough.

i guess what i am saying is that if you want so show that the derivative of an exponential function is something or other, since as usual we learn math backwards, we cannot start by defining our function in terms of integrals. as usual the more you know the easier things are.
as i recall spivak for example doesn't even mention it until he defines
\[\exp=\log^{-1}\] but most texts have you finding derivatives of exponentials right away.

Answer 19

@ta if you are still here, it is usually the case that when you have a variable in the exponent, you have a choice. either
a) first take the log, then rewrite is with the variable on the ground floor , as in: if
\[f(x)=b^x\] then
\[\ln(f(x))=x\ln(b)\]
do some work, and then go back by "exponentiating"
or b) rewrite as
\[b^x=e^{x\ln(b)}\] all the work will be the same in either case

Answer 20

No of course. But I was trying to give the careful answer to "why is e^x special"? Well, it's defined to be special, so in a way its specialness is trivial.

But by god it is useful!

Answer 21

@james yes you are right, so it is not really a miracle since we made it that way!

Answer 22

Yeah, I'm still I just enjoying reading both input on this problem, Thanks!

Answer 23

@sat73 very happy you quoted Spivak. That was my first year text -- love it!

Answer 24

oops! meant I'm still here and enjoy both of ya'll input.