# Derivative of a^x

1. Jun 18, 2014

I can't use the template here.
Find the derivative of $a^x$
I know that it will be $In(a).a^x$

Now, I watched a lecture just now. How he derived this is as follows:
$$\frac{\text{d}}{\text{d}x}a^x=a^x.\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$$

(I omitted some steps). Then he made $\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$ a function defined as $M(a)= \lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$

Do $\frac{\text{d}}{\text{d}x}a^x=a^x.M(a)$ .Now when we make x=0, the derivative of $a^x$ is equal to $M(a)$.
So $M(a)$ can be defined at the slope of the graph of $a^x$ when x=0.

Then he introduced a new variable $e$ so that $M(e)=1$
That means $\frac{\text{d}}{\text{d}x}e^x=e^x$
But we don't yet know what $e$ is. We just defined it.
Now, to find the derivative of $a^x$ what he did was:
Convert it to base $e$. $a^x=e^{In(a)x}$

I don't understand anything after this. Why did we define $M(e)=1$? Why not $M(e)=2$????
And why $e$? Why not $z$?

2. Jun 18, 2014

### micromass

Staff Emeritus
This are questions from a lecture, it is fine to post this in the main math forums. I'll move it to there.

It's an L and not an I, so it's ln, not In. The ln is from logarithmus naturalis.

These are just definitions. There is nothing to understand. He cold have chosen $M(e) = 2$ too, the theory would work perfectly with that choice. But historically, people have always taken $M(e) = 1$ and it's probably not a good idea to break with this history.
One could also argue that we take $M(e) = 1$ because $1$ is simpler than $2$.

We could call it $z$ if we wanted to. It doesn't matter, it's just a name. But again, historically, we have give it the name $e$. So we want to keep this tradition.

3. Jun 18, 2014

What if we took $M(e)=0$? I doubt we can define it like that.

Isn't $e$ the eulers number? This came from the study of compound interest? What does compound interest have to do with this $e$ we just defined?

4. Jun 18, 2014

### micromass

Staff Emeritus
There is certainly some number $a$ such that $M(a) = 0$. In fact, this number is $1$. So $M(1) = 0$.
You can make any definition you like. The question is what we want to do with the definition. We have found that $\frac{d}{dx} a^x = M(a)a^x$. This is true for all $a$. It makes sense to see for which $a$ this equation is 'simplest', whatever that means. Well, you certainly can't argue that the equation takes a simple form if $M(a) = 1$. So that looks promising. And indeed, the choice of $a$ such that $M(a) = 1$ turns out to be such a good choice that we have given it a special name, $e$. Of course, this is posteriori knowledge. A priori, $e$ is just some number that makes the equation $\frac{d}{dx} a^x = M(a)a^x$ simple. It's only after figuring out more math that we really see its importance.

Yes, it is in fact the same number. The relation to compound interest requires a proof that is not easy right now. I'm sure you will do the proof later.
Although $e$ was first discovered with relation to this interest, its real importance is because of its uses in calculus.

5. Jun 18, 2014

Ok, I understand now. I have another question tho!
How can we make $\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$ a function?
Let's leave it and have another limit which is $\lim_{\Delta x \to 0}2+\Delta x$
Now this eventually becomes 2. How can we say that $M(a)=\lim_{\Delta x \to 0}2+\Delta x$ when $\lim_{\Delta x \to 0}2+\Delta x$ is always 2?

6. Jun 18, 2014

### D H

Staff Emeritus
Hold on a second. Was one of those omitted steps the definition of $a^x$?

That choice leads to the *unique* function $\exp(x)$ such that $\frac {d\exp(x)}{dx} = \exp(x)$ and $\exp(0)=1$.

However, the approach taken begs the question, what does $a^x$ mean? A better way to look at it, at least to me, is to define the function $\exp(x)$ as satisfying $\frac {d\exp(x)}{dx} = \exp(x)$ and $\exp(0)=1$ and then proceed from there.

7. Jun 18, 2014

### D H

Staff Emeritus
Every value of $a>0$ results in a specific value of that limit. The mapping from positive $a$ to values of the limit: That's a function.

8. Jun 18, 2014

No. He just used the definition of derivative.

9. Jun 18, 2014

### D H

Staff Emeritus
How did he define $a^x$ then? He had to use some definition to be able to take that step from $a^{x+\Delta x}-a^x = a^x\bigl(a^{\Delta x}-1\bigr)$.

10. Jun 18, 2014

Yeah, he said that $a^x$ is constant and he took it out of the limit.
He said that $\Delta x$ is the thing that is moving, not x.
I don't think I understood this very well

11. Jun 18, 2014

### D H

Staff Emeritus
That step involves using $a^{x+\Delta x} = a^x a^{\Delta x}$. How did he justify that step?

12. Jun 18, 2014

### micromass

Staff Emeritus
You should probably get the book "First course in calculus" by Lang because he uses the same kind of argument to find the derivative of $a^x$.

13. Jun 18, 2014

What he did is this:
Using the definition of derivative,
$$\frac{\text{d}}{\text{d}x}a^x=\lim_{\Delta x \to 0}\frac{a^{x+ \Delta x}-a^{\Delta x}}{\Delta x}$$
Then using the laws of indices, this becomes:
$$\lim_{\Delta x \to 0}\frac{a^xa^{\Delta x}-a^{\Delta x}}{\Delta x}$$
Then,he factorised the numerator and got $a^x(a^{\Delta x}-1)$

Edit:
I think you have told me that at least 10x :shy:
I will buy it but my father is not at home these days. That's why I am watching the lectures first.

14. Jun 20, 2014

### DrewD

He is probably working from the knowledge that the students have from precalculus. It is not rigorous, but the students have already seen the rules for $a^x$ when $a>0$ and $x\in\mathbb{Q}$. I follow this general method in my AP class. We are answering the question, "assuming that $a^x$ is differentiable, what is the derivative?" The next step, after finding that the derivative is proportional to $a^x$ itself, is to wonder if there might be a value of $a$ such that $\frac{d}{dx}a^x=a^x$.

It seems reasonable in an introductory class to assume that $a^x$ follows the same rules for all $a\in\mathbb{R}$ and that the limit in question exists. I always state that these are assumptions that should be dealt with at some point, but not in my class.