# Thread: Differentiation using first principles

1. ## Differentiation using first principles

Hi

How do you differentiate using first principles

eg 3^x

Any help will be appreciated

2. Let $\displaystyle y = f(x) = 3^x$

Then $\displaystyle \frac{dy}{dx}=\lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to\0}\frac{3^{x+h}-3^x}{h}$

$\displaystyle = \lim_{h\to\0}\frac{3^x3^h-3^x}{h} = 3^x \lim_{h\to\0}\frac{3^h-1}{h} = 3^x\ln3$

3. Originally Posted by Sambit
Let $\displaystyle y = f(x) = 3^x$

Then $\displaystyle \frac{dy}{dx}=\lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to\0}\frac{3^{x+h}-3^x}{h}$

$\displaystyle = \lim_{h\to\0}\frac{3^x3^h-3^x}{h} = 3^x \lim_{h\to\0}\frac{3^h-1}{h} = 3^x\ln3$

While true is:

$\displaystyle \lim_{h\to\0}\frac{3^h-1}{h} = \ln(3)$

common knowledge?

4. Originally Posted by nikki1234
Hi

How do you differentiate using first principles

eg 3^x

Any help will be appreciated
Did you just pluck this example out of the air or where you specifically given it and asked to differentiate it from first principles? (Or were you just asked to differentiate it and there was no mention of first principles?)

5. Originally Posted by CaptainBlack
While true is:

$\displaystyle \lim_{h\to\0}\frac{3^h-1}{h} = \ln(3)$

common knowledge?
Derivative of $\displaystyle 3^x$ is always $\displaystyle 3^x\ln3$. Where exactly do you find my mistake?

6. Originally Posted by Sambit
Derivative of $\displaystyle 3^x$ is always $\displaystyle 3^x\ln3$. Where exactly do you find my mistake?
The point being made is that if one already knows that what is there to prove? In other words, have you not used what is to be proved in the proof? Is that really common knowledge?

7. Originally Posted by Plato
The point being made is that if one already knows that what is there to prove? In other words, have you not used what is to be proved in the proof? Is that really common knowledge?
I'd still like to see the response of the OP to post #4. (Many students don't realise that the questions they get given in their textbook etc. are sanitised and cooked to work. A lot of time gets wasetd when it turns out that the posted question has just been plucked out of the air by the student).

8. I can't really follow. In my calculus book, derivative of $\displaystyle 3^x$ is calculated from first principle; and it is done in the way I did it here. Is it not right that every algebraic function can be differentiated using first principle?

9. Originally Posted by Sambit
I can't really follow. In my calculus book, derivative of $\displaystyle 3^x$ is calculated from first principle; and it is done in the way I did it here. Is it not right that every algebraic function can be differentiated using first principle?
What they are asking is "How do you know that this limit"

$\displaystyle \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)$

You can't use L'hosptials rule because we are trying to prove how to take the derivative.
It is the same for using a Taylor series.

10. Originally Posted by TheEmptySet
What they are asking is "How do you know that this limit"

$\displaystyle \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)$
..which is equivalent to what you needed to prove in the first place, it is the derivative of the given function at 0.

CB

11. Ok; so you guys are saying I need to prove $\displaystyle \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)$ first?

12. Originally Posted by Sambit
Ok; so you guys are saying I need to prove $\displaystyle \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)$ first?
Here is a suggestion: $\displaystyle \frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}$

13. Originally Posted by Plato
Here is a suggestion: $\displaystyle \frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}$
I know the process. i just thought it would not be needed.. Here it goes:

$\displaystyle \frac{3^h-1}{h}=\frac{e^z-1}{\frac{z}{\ln 3}}$ (where $\displaystyle z=h\ln 3$)
$\displaystyle =\ln 3\lim_{z\to\0}\frac{e^z-1}{z}=\ln 3$

But now if you say why is $\displaystyle \lim_{z\to\0}\frac{e^z-1}{z}=1$? , I can't answer because I have learned it as a result without proof.

14. Originally Posted by Sambit
But now if you say why is $\displaystyle \lim_{z\to\0}\frac{e^z-1}{z}=1$? , I can't answer because I have learned it as a result without proof.
There are multiple ways that textbook authors present this material. I like the way it is done in Gillman & McDowell.
But theirs is much different from your question, which is the way James Stewart’s books do it.
Stewart simply defines $\displaystyle \mathbf{e}$ to the number such that
$\displaystyle \displaystyle\lim _{h \to 0} \frac{{e^h - 1}}{h} = 1$.

15. Originally Posted by Sambit
I know the process. i just thought it would not be needed.. Here it goes:

$\displaystyle \frac{3^h-1}{h}=\frac{e^z-1}{\frac{z}{\ln 3}}$ (where $\displaystyle z=h\ln 3$)
$\displaystyle =\ln 3\lim_{z\to\0}\frac{e^z-1}{z}=\ln 3$

But now if you say why is $\displaystyle \lim_{z\to\0}\frac{e^z-1}{z}=1$? , I can't answer because I have learned it as a result without proof.
Then it can't be used in a derivation from first principles

CB

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