Well, it's all in the title.
Any help is appreciated!
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.
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 $
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.
$\displaystyle \displaystyle y &= a^x \\ \ln{(y)} &= \ln{\left(a^x\right)} \\ \ln{(y)} &= x\ln{(a)} \\ \frac{d}{dx}\left[\ln{(y)}\right] &= \frac{d}{dx}\left[x\ln{(a)}\right] \\ \frac{d}{dy}\left[\ln{(y)}\right]\,\frac{dy}{dx} &= \ln{(a)} \\ \frac{1}{y}\,\frac{dy}{dx} &= \ln{(a)} \\ \frac{dy}{dx} &= y\ln{(a)} \\ \frac{dy}{dx} &= a^x\ln{(a)}$
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.
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.
Doesn't that mean that the limit does not exist?
What am I missing.