Unclear Proving the Product Derivative rule using the definition of a derivative

Question

I am trying to prove the product rule derivative using the definition of the derivative.

Definition of derivative --> $h'(c)= \lim_{ n \to 0} = \frac{h(c+n) - h(c)}{n}$

Prove $h'(c) = f'(c)g(c) + f(c)g'(c) $

$h'(c)= \lim_{ n \to 0} = \frac{h(c+n) - h(c)}{n} = \lim_{n \to 0} \frac{f(c+n)g(c+n) - f(c)g(c)}{n}$

$\lim_{n \to 0} \frac{(f(c+n)g(c+n) - f(c+n)g(c)) - (f(c+n)g(c)- f(c)g(c))}{n}$

$\lim_{n \to 0} f(c+n)( \frac{g(c+n)-g(c)}{n}) + \lim_{n \to 0} g(c) \frac{f(c+n)+f(c)}{n}$

From the above its easy to see that it is equals to - $h'(c) = f'(c)g(c) + f(c)g'(c) $

But I have major confusions on:

$\lim_{n \to 0} \frac{(f(c+n)g(c+n) - f(c+n)g(c)) - (f(c+n)g(c)- f(c)g(c))}{n}$

How did this happen from $\lim_{n \to 0} \frac{f(c+n)g(c+n) - f(c)g(c)}{n}$

It seems you have mixed together the product and chain rules. They are quite different. — coffeemath, Sep 13 '22 at 10:43
Does this answer your question? Derivative of a product and derivative of quotient of functions theorem: I don't understand its proof — Anne Bauval, Sep 13 '22 at 10:48

score 2 · Answer 1 · answered Sep 13 '22 at 10:45

2

There is a typo when you write $\frac{(f(c+n)g(c+n) - f(c+n)g(c)) - (f(c+n)g(c)- f(c)g(c))}{n}$. You have

$$ \frac{(f(c+n)g(c+n) - f(c)g(c))}{n} = \frac{(f(c+n)g(c+n) - f(c+n)g(c)) + (f(c+n)g(c)- f(c)g(c))}{n}$$

answered Sep 13 '22 at 10:45

how did you get from the 2nd to 3rd line? it does not make any sense to me – user307640 Sep 13 '22 at 15:44
@user307640 In the third line you have $- f(c+n)g(c)) + f(c+n)g(c) = 0$. – Kritiker der Elche Sep 16 '22 at 23:21

1 Answers1