# Proving the Product Rule

1. Jul 8, 2013

### Lemniscates

Hello all,

I'm having trouble with proving that the derivative of f(x)*g(x) is f'(x)*g(x)+f(x)*g'(x).
Now, I've already seen the actual proof, and I can understand its reasoning, but the first time I tried to prove without looking at the solution, this is what I wrote before I became rather confused:

So, using the limit definition of the derivative, I get that the derivative of f(x)*g(x) is:

$\displaystyle\lim_{\Delta x \rightarrow 0} {\frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)} {\Delta x}}$

I can rewrite this as:
$\displaystyle\lim_{\Delta x \rightarrow 0} {(f(x+\Delta x)\frac{g(x+\Delta x) }{\Delta x} - f(x)\frac{g(x)}{\Delta x})}$

Then I used some limit rules (multiplication and subtraction of limits) to rewrite the limit:
$\displaystyle\lim_{\Delta x \rightarrow 0} {f(x+\Delta x)} \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x+\Delta x)}{\Delta x}} - \displaystyle\lim_{\Delta x \rightarrow 0} {f(x)} \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x)}{\Delta x}}$

I believe that, because f(x) and g(x) are differentiable (and therefore continuous), this evaluates to:
$f(x) \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x+\Delta x)}{\Delta x}} - f(x) \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x)}{\Delta x}}$

Then I can factor out f(x) and be left with :
$f(x)( \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x+\Delta x)}{\Delta x}} - \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x)}{\Delta x}})$

Then, because the the limit of a difference is the same as the difference of the two terms' limits:

$f(x)( \displaystyle\lim_{\Delta x \rightarrow 0}{\frac{g(x+\Delta x)-g(x)}{\Delta x}})$

I finally get: $f(x)g'(x)$

I realize that this isn't the right answer. Where am I going wrong?

Last edited: Jul 8, 2013
2. Jul 8, 2013

### Infrared

You can only write the limit of a difference as a difference of two limits if the two limits actually exist. The two limits you have written here will not exist in general.

3. Jul 8, 2013

### mathman

f(x+h)g(x+h) - f(x)g(x) = {f(x+h)g(x+h) - f(x)g(x+h)} + {f(x)g(x+h) - f(x)g(x)}

Let h -> 0. Assume f and g are well behaved.
{f(x+h)g(x+h) - f(x)g(x+h)}/h -> g(x)f'(x)
{f(x)g(x+h) - f(x)g(x)}/h -> f(x)g'(x)

I hope this clarifies it.

4. Jul 8, 2013

### Lemniscates

Thank you, HS, for helping me find the hole in my proof.

And thank you mathman, but I've already seen the actual proof, which makes sense to me. I just needed someone to explain why my version didn't work.

5. Jul 8, 2013

### CubicFlunky77

Why not use the chain rule for multiple variables employing parametric identities?

Let $W=uv$

where $u = x(t)$ and $v=y(t)$

then $dW=W_udu + W_vdv$

Using approximations we can say: $\frac{dW}{dt} = \frac{\partial W}{\partial u} \frac{du}{dt} + \frac{\partial W}{\partial v} \frac{dv}{dt}$

Because of the parametric equations we can define the partial terms and $u$ and $v$ with respect to each other and to $t$;

Thus, $W_u=v , \frac{du}{dt}=x'(t) , W_v=u , \frac{dv}{dt}=y'(t)$

Then just plug-and-play! :

$y(t)x'(t) + x(t)y'(t)$

To show it using limits simply carry over the epsilon/delta definition for single variables to multiple variables (partials).

6. Jul 8, 2013

### WannabeNewton

That has to be one of the most needlessly convoluted way to prove something I have personally ever seen. The standard proof of the single-variable product rule using single-variable techniques is in and of itself simpler and way more minimalist.

7. Jul 8, 2013

### micromass

Staff Emeritus
Yes, because the OP surely has seen the chain rule for multivariable calculus before if he's trying to prove the product rule.

Please don't give out responses that are above the current level of understanding of the OP. Thanks.

8. Jul 9, 2013

### willem2

The hole is that you replace $f(x+ \Delta x )$ by $f(x )$ between the third and the fourth expression, where you said
What HS-Scientist wrote isn't really a problem, because you could have kept the entire expression as a single limit anyway.

9. Jul 9, 2013

### lurflurf

This thread is hysterical. I think it is worthwhile to learn both proofs. The Newton quotient proof is very visual we note (perhaps by drawing a rectangle) that

Δ(fg)=(Δf)g+f(Δg)+Δ(f)(Δg)

Many people like the nonsymmetrical variation for some reason.
HS-Scientist's objection is dealt with by supposing f' and g' exist

WannabeNewton that proof is extremely straight forward how is it needlessly convoluted?
The distinction between one and several variables is entirely artificial.
single variable calculus exercise
$$\dfrac{d}{dx} c \, x=c$$
several variable calculus exercise
$$\dfrac{\partial}{\partial x} z \, x=z$$
exactly the same idea
The OP claims knowledge of the chain rule. He/She therefore knows the multivariable chainrule with or without being aware of it.
The second proof has several benefits. It helps one develop derivative intuition. Many absurd errors are commited because students reading crap books like Stewart have no intuition. To them (uv)'=u'+v' or (x^x)'=x^x is perfectly reasonable. The key concept is the derivative of a function where the variable appears multiple times can be found by adding the derivatives of each appearance. This is because differentiation is linear.

10. Jul 9, 2013

### CubicFlunky77

The cynicism and hostility was uncalled for. Proving the product rule for single variables is a very typical homework problem in the vast majority of AP Calculus high-school classes let alone colleges. Since this is not the homework section I had no idea the OP didn't know the chain-rule for multiple variables. Also, I personally struggled to understand the product rule proof for single variables. The method I used, was done in my community college class and is 100% crystal clear to me.

11. Jul 9, 2013

### lurflurf

^Bravo

12. Jul 9, 2013

### micromass

Staff Emeritus
Now you're just guessing. There is no reason to assume the OP had any acquaintance with multivariable calc. In fact, if somebody asks something about single-variable calculus, then it's simply common sense to not start using tools that are not covered in a single-variable calculus course.

Anyway, let's keep this thread on topic.

13. Jul 9, 2013

### lurflurf

That's not guessing, it is being open to possibility. It is not like the OP listed the topics known. In any case the two topics are within a year of each other anyway, and topics are presented in order in different books/classes. One might read a book out of order. Most importantly the most common ordering of calculus topics is nonoptimal. If one has the misfortune of a book that does the product rule before the chain rule, one may simply flip to the chain rule bit read it then flip back, problem solved.

The original mistake is ambiguous.
On second reading HS-Scientist's comment is spot on if the mistake started there.
The OP's proof is quite correct (reasoning and result) in the case g(x)=0
(also the case f'(x)=0)
When this is not the case HS-Scientist's is correct the limit manipulation is invalid.
The best thing to do as I said above is write
f(x+h)=f(x)+(f(x+h)-f(x))=
g(x+h)=g(x)+(g(x+h)-g(x))=

(f(x+h)g(x+h)-f(x)g(x))/h=(f(x+h)-f(x))g(x)/h+f(x)(g(x+h)-g(x))/h
as OP omits the term (f(x+h)-f(x))g(x)/h it is assumed that
lim (f(x+h)-f(x))g(x)/h=0

either f'(x)=0 or g(x)=0
which is not true in general

14. Jul 9, 2013

### micromass

Staff Emeritus
Right. And in that case it's best to assume a minimal amount of knowledge. Assuming a knowledge of multivariable calculus is not ok here.

Do you have any book that proves the multivariable chain rule before the single variable product rule?? Do you really think it's misfortune to prove the product rule first? Do you really think we should let students wait a year to actually prove the product rule when a very simple limit argument suffices? With all due respect, but what you're proposing is not in the best interest of the students.

15. Jul 9, 2013

### MarneMath

I don't see what the big deal is. OP posted a question, OP received answer, user posted an alternate method. I think being aware of how to prove a proof different ways doesn't hurt. :)

16. Jul 9, 2013

### micromass

Staff Emeritus
Let's get back to the math now and keep things on-topic.

17. Jul 9, 2013

### Infrared

No, this is perfectly valid if you are allowed to break up limits. If f is continuous (which you know since we are assuming that f is differentiable, then $lim_{\Delta x \rightarrow 0} f(x+\Delta x)=f(x)$.

How would you be able to do the same proof without breaking up the limit?