Differentiation using first principles

**nikki1234** · May 6th 2011, 11:39 AM

Hi

How do you differentiate using first principles

eg 3^x

Any help will be appreciated

**Sambit** · May 6th 2011, 11:04 PM

Let $y = f(x) = 3^x$

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

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

**CaptainBlack** · May 7th 2011, 09:58 PM

Originally Posted by Sambit

Let $y = f(x) = 3^x$

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

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

While true is:

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

common knowledge?

**mr fantastic** · May 7th 2011, 11:22 PM

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?)

**Sambit** · May 8th 2011, 06:25 AM

Originally Posted by CaptainBlack

While true is:

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

common knowledge?

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

**Plato** · May 8th 2011, 06:31 AM

Originally Posted by Sambit

Derivative of $3^x$ is always $3^x\ln3$ . 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?

**mr fantastic** · May 8th 2011, 12:12 PM

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).

**Sambit** · May 8th 2011, 09:23 PM

I can't really follow. In my calculus book, derivative of $3^x$ 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?

**TheEmptySet** · May 8th 2011, 09:27 PM

Originally Posted by Sambit

I can't really follow. In my calculus book, derivative of $3^x$ 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"

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

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.

**CaptainBlack** · May 8th 2011, 10:49 PM

Originally Posted by TheEmptySet

What they are asking is "How do you know that this limit"

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

..which is equivalent to what you needed to prove in the first place, it is the derivative of the given function at 0.

CB

**Sambit** · May 9th 2011, 04:48 AM

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

**Plato** · May 9th 2011, 05:02 AM

Originally Posted by Sambit

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

Here is a suggestion: $\frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}$

**Sambit** · May 9th 2011, 07:32 AM

Originally Posted by Plato

Here is a suggestion: $\frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}$

I know the process. i just thought it would not be needed.. Here it goes:

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

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

**Plato** · May 9th 2011, 07:49 AM

Originally Posted by Sambit

But now if you say why is $\lim_{z\to\0}\frac{e^z-1}{z}=1$ ? , 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 $\mathbf{e}$ to the number such that
$\displaystyle\lim _{h \to 0} \frac{{e^h - 1}}{h} = 1$ .

**CaptainBlack** · May 9th 2011, 07:51 AM

Originally Posted by Sambit

I know the process. i just thought it would not be needed.. Here it goes:

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

But now if you say why is $\lim_{z\to\0}\frac{e^z-1}{z}=1$ ? , 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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