# Differentation of a^x

• Aug 6th 2009, 06:22 PM
chengbin
Differentation of a^x
I don't get how the derivative of the exponential function $a^x$ is derived.

$\frac {dy}{dx}=\lim_{h \to 0} \frac {a^{x+h}-a^x}{h}$

$=\lim_{h \to 0} \frac {a^x * a^h-a^x}{h}$

$=a^x \lim_{h \to 0} \frac {a^h-1}{h}$

$=a^x \lim_{h \to 0} \frac {a^h-a^0}{h}$
This is the point where I start to not get.

$=a^xf'(0)$

(why does $\frac {a^h-a^0}{h}$ = $f'(0)$)

The value of $f'(0)$ is the derivative of $y=f(x)=a^x$ at $x=0$

Assuming that $e$ is the value of $a$ at which the slope of the tangent line at $f(0)$ is 1. (Why assume e?)

I get the rest.

I got a question for the differentation method of $\ln x$ as well.

Since $\lim_{t \to \infty} \bigg ( 1+\frac {1}{t}\bigg )^t=e$

$\lim_{k \to 0}\ln(1+k)^\frac {1}{k}=1$ (how?)
• Aug 6th 2009, 07:26 PM
AlephZero
Quote:

Originally Posted by chengbin
$=a^x \lim_{n \to \infty} \frac {a^h-a^0}{h}$
This is the point where I start to not get.

$=a^xf'(0)$

(why does $\frac {a^h-a^0}{h}$ = $f'(0)$)

The value of $f'(0)$ is the derivative of $y=f(x)=a^x$ at $x=0$

Assuming that $e$ is the value of $a$ at which the slope of the tangent line at $f(0)$ is 1. (Why assume e?)

I'm assuming your $n$'s are $h$'s in the limit; in other words, that it's just a typo? Anyway, since by definition,

$f^\prime (0)=\lim_{h\to\infty} \frac{f(0+h)-f(0)}{h}=\lim_{h\to\infty}\frac{a^{0+h}-a^0}{h}=\lim_{h\to\infty}\frac{a^h-1}{h},$

• Aug 6th 2009, 07:40 PM
AlephZero
Here's a better proof, just for giggles:

$(a^x)^\prime=[(e^{\ln a})^x]^\prime$

$= [e^{(\ln a)x}]^\prime =e^{(\ln a)x}[(\ln a)x]^\prime$

$= (e^{\ln a})^x(\ln a)=a^x\ln a.$
• Aug 7th 2009, 05:26 AM
chengbin
Quote:

Originally Posted by AlephZero
Here's a better proof, just for giggles:

$(a^x)^\prime=[(e^{\ln a})^x]^\prime$

$= [e^{(\ln a)x}]^\prime =e^{(\ln a)x}[(\ln a)x]^\prime$

$= (e^{\ln a})^x(\ln a)=a^x\ln a.$

That's when you know the formula $a'^{x}=a^x\ln a$

Oh BTW, those were typos, they're fixed now.
• Aug 7th 2009, 08:07 AM
e^(i*pi)
For constant a:

Let $y=a^x$

Taking the log of both sides gives $ln(y) = x\, ln(a)$

Using implicit differentiation: $\frac{1}{y} \, \frac{dy}{dx} = ln(a)$

Isolate dy/dx: $\frac{dy}{dx} = y\, ln(a)$

Recall step 1: $\frac{dy}{dx} = a^x\, ln(a)$
• Aug 7th 2009, 08:10 AM
AlephZero
Quote:

Originally Posted by chengbin
That's when you know the formula $a'^{x}=a^x\ln a$

Oh BTW, those were typos, they're fixed now.

Actually, no, that's a derivation of the formula $(a^x)^\prime=a^x\ln a.$ All it requires is knowledge of how to differentiate $e^x.$
• Aug 7th 2009, 10:33 AM
chengbin
Guys, there is a reason I typed the "old fashioned" way of differentiating $y=a^x$. I'm not looking for proofs of the formula, or other ways to solve it. I'm just looking to have my questions in brackets beside the steps answered (and the limit question on the bottom).
• Aug 7th 2009, 11:47 AM
Plato
Quote:

Originally Posted by chengbin
Guys, there is a reason I typed the "old fashioned" way of differentiating $y=a^x$. I'm not looking for proofs of the formula, or other ways to solve it. I'm just looking to have my questions in brackets beside the steps answered (and the limit question on the bottom).

In your OP the limit is incorrect.
$f'\left( {x_0 } \right) = \lim _{h \to {\color{red}0}} \left( {\frac{{f\left( {x_0 + h} \right) - f\left( {x_0 } \right)}}
{h}} \right)$

Thus, we end up with $f'\left( x \right) = a^x \lim _{h \to 0} \left( {\frac{{a^h - 1}}{h}} \right)$.

The real problem at that point is to know (or show if need be) that $\lim _{h \to 0} \left( {\frac{{a^h - 1}}{h}} \right) = \ln (a)$

If that must be shown it is difficult. From what you have written it looks like you are trying to follow the proof in James Stewart's textbook. That is one of the best proof's of this.

Notice that he defines $e$ as the number such that $\lim _{h \to 0} \left( {\frac{{e^h - 1}}{h}} \right) = 1$.

Does that help at all?
• Aug 7th 2009, 12:00 PM
chengbin
Ahhh, I keep making typos.

So it seems Kumon copied James Stewart's textbook? I simply copied the sheet's guide.
• Aug 7th 2009, 12:04 PM
Plato
Quote:

Originally Posted by chengbin
So it seems Kumon copied James Stewart's textbook? I simply copied the sheet's guide.

Well, in Stewart’s book that discussion takes about five pages to fully develop.
• Aug 7th 2009, 01:58 PM
chengbin
Quote:

Originally Posted by Plato
Well, in Stewart’s book that discussion takes about five pages to fully develop.

It went on in Kumon as well (only for another page), but I didn't type it because I get it.

Any web link available for me to see the discussion?

EDIT: nevermind, I found it on Google Books. It is the same thing as Kumon, except explained in more depth, with graphs (Kumon "magically" expects you to understand this, or they expect you to spend time in Kumon with the professor so they can shrink the number of worksheets)
• Aug 7th 2009, 02:59 PM
Plato
Quote:

Originally Posted by chengbin
It went on in Kumon as well (only for another page), EDIT: nevermind, I found it on Google Books. It is the same thing as Kumon, except explained in more depth, with graphs (Kumon "magically" expects you to understand this, or they expect you to spend time in Kumon with the professor so they can shrink the number of worksheets)

What is "Kumon"?
• Aug 7th 2009, 03:47 PM
skeeter
Quote:

Originally Posted by Plato
What is "Kumon"?

supposedly, another "silver bullet" educational method to slay the dragon of innumeracy in grades K-12.
• Aug 7th 2009, 04:10 PM
chengbin
Quote:

Originally Posted by Plato
What is "Kumon"?

A math (and reading, but known for math) tutoring program that's known to teach little kids basic elementary math but also has material all the way to calculus, linear algebra, and statistics (but almost nobody makes it there. One word sums up Kumon's curriculum below calculus, brutal)
• Aug 7th 2009, 04:39 PM
chengbin
Quote:

Originally Posted by skeeter
supposedly, another "silver bullet" educational method to slay the dragon of innumeracy in grades K-12.

Unforunately it is not. The content taught below calculus is way too difficult. The way they taught it is kind of "minimalistic" (thus requiring a really bright mind to understand if no one helped you, even though it is a "self taught" program). It makes perfect sense when you get it and look back, but not when you first learn it. It also goes into ridiculous depth (way more than school) that you won't ever need (factorization and optimization being the worst offenders).

The good thing is the university level topics are not explored in ridiculous depth. I guess those are not for catching up school, but more for curious minds (like me) to get an idea of the material taught in university, and ease your way. For example, Kumon doesn't teach multi variable calculus, or inverse trigonometry. Linear algebra section only teaches matrix, mapping, and transformations.

And for those reasons, It is quite rare to find a student that finishes Kumon. Most people quit after a while because they get too frustrated either doing the overly challenging work, or frustrated from not understanding.