Question 1

Is there reasons why there are many proofs?

Accepted Answer

There are several proofs because each statement or concept needs to be proven individualy in order to be accepted. Proofs can also bundle up and be used to prove an even higher concept.

Question 2

According to the Limit rules lim x->a f(x).g(x)  = lim x->a  f(x) . lim x->a  g(x) . But why when the derivative is calculated this rule is violated?

Accepted Answer

Interesting question!  The derivative of a function is not just the limit of the function, but rather is a limit of something more complicated that is related to the function.

Note that the derivative of f(x)g(x) is limit h-->0 of [f(x+h)g(x+h) - f(x)g(x)]/h,
the derivative of f(x) is  limit h-->0 of [f(x+h) - f(x)]/h, 
and the derivative of g(x) is limit h-->0 of [g(x+h) - g(x)]/h.

In general, [f(x+h)g(x+h) - f(x)g(x)]/h is not the product of [f(x+h) - f(x)]/h and [g(x+h) - g(x)]/h, so we can't just use the product property of limits to conclude that the derivative of f(x)g(x) is the product of the derivatives of f(x) and g(x).

Have a blessed, wonderful day!

Question 3

At 1:00 I did not understand why d/dx[f(x)*g(x)] =

lim   [f(x+h)g(x+h) - f(x)g(x)]/h 
h->0

Accepted Answer

This is the definition of the derivative with f(x)·g(x) substituted in for f(x).

Question 4

Why does lim h->0 f(x + h) is equal to f(x) ?

Accepted Answer

Because if a function is continuous, then the limit of the function is equal to the function of the limit. That is,
lim_{x-> c} f(x)=f(lim_{x->c} x)=f(c)

In this case, we have assumed f to be differentiable, and therefore continuous, so lim_{h->0} f(x+h)=f(x+0)=f(x).

Question 5

How does Sal go from step 2 to step 3 (minute 5:47 to 6:26)? I mean, how is it concluded that lim h->0 [f(x+h)*(g(x+h)-g(x))/h] = lim h->0 [f(x+h)]*lim h->0 [(g(x+h)-g(x))/h]?

In other words, if the limit of two functions multiplying each other is lim h->0 [f(x)g(x)] = f’(x)g(x) + f(x)g’(x), then why does Sal says that “the limit if the products is the product of the limits”? Wouldn’t we fall into a redundance in the proof? Please help!

Accepted Answer

You're confusing the product rule for derivatives with the product rule for limits. The limit as h->0 of f(x)g(x) is
[lim f(x)][lim g(x)], provided all three limits exist. f and g don't even need to _have_ derivatives for this to be true.

The derivative of f(x)g(x) if f'(x)g(x)+f(x)g'(x). This is not the same as the limit.

Question 6

at 1:50, did we assume the two functions be on a same graph or something? i did not get the step... why do we multiply (f.g)(x+h) and subract it by f.g(x) and have only one h in the denominator? how to visualise two functions of distinct graphs, f(x) and g(x) being multiplied, that too of their slopes?

thanks

Accepted Answer

Let the function H(x) = f(x)*g(x). H(x) can be plotted on the same graph as f(x) and g(x) because all 3 of them are functions of "x" and only "x".

We have to find the derivative of H(x), which is 
_lim[h-->0] (H(x+h) - H(x))/h_

H(x+h) is f(x+h)g(x+h) and H(x) is f(x)g(x), the limit can be written as
_lim[h-->0] (f(x+h)*g(x+h) - f(x)*g(x))/h_

Question 7

According to the Limit rules lim x->a f(x).g(x) = lim x->a f(x) . lim x->a g(x) . But why when the derivative is calculated this rule is violated?

Accepted Answer

The product limit rule is not violated.

(f(x)g(x))' = lim h->0 (f(x+h)g(x+h) - f(x)g(x)) / h

We cannot split the limit into the product of lim h->0 (f(x+h) - f(x))/h and lim h->0 (g(x+h) - g(x))/h.

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 2

Proving the product rule

Without further ado, we present to you the proof!

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