How to show rule of product differentiation

Question

I have to show that

$$(fg)' = f'g + fg'$$

How would I go about this? I have to use that limit right, but what then?

Thanks for the help

http://math.stackexchange.com/questions/497602/product-rule-intuition/1950282#1950282 — , Feb 26 '17 at 14:03

score 3 · Answer 1 · answered Feb 26 '17 at 13:16

3

Note that $$\frac{f(x+h)g(x+h)-f(x)g(x)}h=\frac{f(x+h)-f(x)}hg(x+h)+f(x)\frac{g(x+h)-g(x)}h$$

answered Feb 26 '17 at 13:16

Hagen von Eitzen

382,479

score 2 · Accepted Answer · answered Feb 26 '17 at 13:17

2

Yes, you have to apply the limit. It comes by noticing that

$$\small f(x+h)g(x+h)-f(x)g(x)=\underbrace{f(x+h)g(x+h)-f(x)g(x+h)}_{f'(x)g(x)}+\underbrace{f(x)g(x+h)-f(x)g(x)}_{f(x)g'(x)}$$

Can you see where to go from here?

answered Feb 26 '17 at 13:17

Simply Beautiful Art

76,603

Either I'm dumb af, or I don't know.. trying to figure this out and I simply don't understand the part after the equal part – BloodDrunk Feb 26 '17 at 13:34
@BloodDrunk Notice the the limit definition of the derivative wants us to take $$\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}h$$And apply my answer to the numerator. You'll notice that $g(x+h)$ and $f(x)$ will factor out, and all that will remain is $$f'(x)g(x)+f(x)g'(x)$$ – Simply Beautiful Art Feb 26 '17 at 13:36

score 0 · Answer 3 · answered Feb 26 '17 at 14:53

There is another way using logarithmic differentiation $$y=f\times g\implies \log(y)=\log(f)+\log(g)$$ Differentiate both sides $$\frac{y'} y=\frac{f'} f+\frac{g'} g$$ $$y'=y \left(\frac{f'} f+\frac{g'} g \right)=f g \left(\frac{f'} f+\frac{g'} g \right)=f'g+fg'$$ You could play the same game for the quotient rule or the power law.

How to show rule of product differentiation

3 Answers3