Question about proving the product rule

Question

I am a little confused by one of the primary steps taken to prove this useful rule.

We know that$$\frac{d}{dx}[f(x)g(x)]= \lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}.$$

This is where i am confused. Looking at this equation there is not much if anything we can do. So how i was taught this proof, we would add
$-g(x+h)f(x)+g(x+h)f(x)$ to the numerator in order to manipulate it, then simplify it to get the product rule, but what confuses me is how $-g(x+h)f(x)+g(x+h)f(x)$ was put into the fraction in the first place, multipication? and also why that specific set was chosen to be added(appart from the fact that adding a sum and difference value wont change the total value of the original equation). Was it because whoever was playing around with this knew it would simplify neatly?

Thanks

Answer 1

Given the derivative of $f(x)\cdot g(x)$, you get this with the limit definition of the derivative as you mentioned:

$$\lim_{h\to0} {f(x+h)g(x+h)-f(x)g(x) \over h}$$

Which you can manipulate by subtracting and adding $f(x+h)g(x)$ because that does not change the overall value of the expression, because subtracting then adding the same thing results in zero:

$$\lim_{h\to0} {f(x+h)g(x+h)-f(x)g(x) \over h} = \lim_{h\to0} {f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x) \over h}$$

And then you can split the single limit into two:

$$\lim_{h\to0} {f(x+h)g(x+h)-f(x+h)g(x) \over h} + \lim_{h\to0} {f(x+h)g(x)-f(x)g(x) \over h}$$

Then factor:

$$\lim_{h\to0} {f(x+h)(g(x+h)-g(x)) \over h} + \lim_{h\to0} {g(x)(f(x+h)-f(x)) \over h}$$

Which then you can rewrite and evaluate:

$$\lim_{h\to0} f(x+h){g(x+h)-g(x) \over h} + \lim_{h\to0} g(x){f(x+h)-f(x) \over h}=f(x)g'(x)+g(x)f'(x)$$

Answer 2

The short answer is yes. There are precedents for this sort of thing. For example

$x^2-y^2 = x^2 -xy + xy - y^2 = x(x-y)+y(x-y)=(x-y)(x+y)$

Answer 3

If you find the trick of add/subtract a bit confusing then you can start from the equations $$f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h},g'(x)=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$$ and find expressions for $f(x+h), g(x+h) $ as $$f(x+h) =f(x) +h(f'(x) +\phi(h)) , \\g(x+h) =g(x) +h(g'(x) +\psi(h)) $$ where $\phi(h), \psi(h) $ tend to $0$ with $h$. Then we can see that $$f(x+h) g(x+h) - f(x) g(x)\\=h\{f(x) (g'(x)+\psi(h)) +(f'(x)+\phi(h)) g(x) \}\\ + h^2(f'(x)+\phi(h))(g'(x)+\psi(h)) $$ Now divide the equation by $h$ and let $h\to 0$ to get $$\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=f(x)g'(x)+f'(x)g(x)$$ as desired.

Another technique is classical one to use the symbol $\Delta$ to denote difference. Thus $$\Delta f=f(x+h) - f(x), \Delta g=g(x+h) - g(x) $$ so that $$f(x+h) =f(x) +\Delta f, g(x+h) =g(x) +\Delta g$$ so that $$\frac{f(x+h) g(x+h) - f(x) g(x)} {h} =f(x) \frac{\Delta g} {h} +g(x) \frac{\Delta f} {h} +h\cdot\frac{\Delta f} {h} \frac{\Delta g} {h} $$ and letting $h\to 0$ we get $$(f(x) g(x)) '=f(x) g' (x) +f'(x) g(x) +0\cdot f'(x) g'(x) $$ as desired.