Derivative of a^x: Learn to Calculate M(a) and e^x

[adjacent]

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 $z$?

 

[micromass]

I can't use the template here.

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

Find the derivative of $a^x$
I know that it will be $In(a).a^x$

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

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 $z$?

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.

 

[adjacent]

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.

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?

 

[micromass]

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

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.

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?

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.

 

[adjacent]

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?

 

[D H]

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).

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

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 $z$?

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.

 

[D H]

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?

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.

 

[adjacent]

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

No. He just used the definition of derivative.

 

[D H]

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

 

[adjacent]

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

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

 

[D H]

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

 

[micromass]

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 $a^x$.

 

[adjacent]

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

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:

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 $a^x$.

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.

 

[DrewD]

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

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.