Well, it's all in the title.

Any help is appreciated!

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- Sep 3rd 2011, 04:33 PMrtblueWhy doesn't the power rule work with exponential functions?
Well, it's all in the title.

Any help is appreciated! - Sep 3rd 2011, 05:10 PMTheChazRe: Why doesn't the power rule work with exponential functions?
No, it's not!

Are you referring to the derivative of a constant power?

If so, why on earth*would they*be the same?!?!

The proof of the power rule (at least for positive integers) relies on the binomial theorem, which doesn't have an obvious connection with an/the exponential function. - Sep 3rd 2011, 06:20 PMSorobanRe: Why doesn't the power rule work with exponential functions?
Hello, rtblue!

The power rule works for a special form: .$\displaystyle f(x) \:=\:x^{\text{constant}}$

The exponential function is a different animal: .$\displaystyle f(x) \:=\:\text{(constant)}^x $

- Sep 3rd 2011, 08:12 PMProve ItRe: Why doesn't the power rule work with exponential functions?
Try differentiating $\displaystyle \displaystyle a^x$ using first principles or some other method you already know of (like the chain rule)...

Maybe have a read of this, then use what you have learnt from here to extend to differentiating $\displaystyle \displaystyle y = a^x$ using the Chain Rule.

$\displaystyle \displaystyle y &= a^x \\ \ln{(y)} &= \ln{\left(a^x\right)} \\ \ln{(y)} &= x\ln{(a)} \\ \frac{d}{dx}\left[\ln{(y)}\right] &= \frac{d}{dx}\left[x\ln{(a)}\right] \\ \frac{d}{dy}\left[\ln{(y)}\right]\,\frac{dy}{dx} &= \ln{(a)} \\ \frac{1}{y}\,\frac{dy}{dx} &= \ln{(a)} \\ \frac{dy}{dx} &= y\ln{(a)} \\ \frac{dy}{dx} &= a^x\ln{(a)}$ - Sep 3rd 2011, 09:05 PMterrorsquidRe: Why doesn't the power rule work with exponential functions?
I just looked over a few people explanations of finding the derivative of an exponential from first principles and I was stuck on one part:

e.g.

$\displaystyle \lim_{h \to\0} \frac{a^{x+h}-a^x}{h}$

$\displaystyle \lim_{h \to\0} \frac{a^x(a^h-1)}{h}$

$\displaystyle \frac{a^x\left(\lim_{h \to\0}a^h-1\right)}{h}$

$\displaystyle \lim_{h \to\0}\frac{0}{h}$

Doesn't that mean that the limit does not exist?

What am I missing. - Sep 3rd 2011, 09:23 PMProve ItRe: Why doesn't the power rule work with exponential functions?
- Sep 4th 2011, 06:25 AMHallsofIvyRe: Why doesn't the power rule work with exponential functions?
This is incorrect. $\displaystyle a^h- 1$ is NOT 0 for all h (or even all small h). It is true that if you take the limits of numerator and denominator separately, you get 0 for

**both**but that is true of**all**derivative limits.

Quote:

Doesn't that mean that the limit does not exist?

What am I missing.