Question regarding Product Rule of Derivatives

Question

According to the Limit principles the $\lim\limits_{x \to a} f(x).g(x)$ is $\lim\limits_{x \to a} f(x).\lim\limits_{x \to a} g(x)$.

And also the definition of Derivatives is also $f'(x)= \lim\limits_{h \to 0} \frac {f(x+h)-f(x)}{h}$

Hence the derivative of $F'(x)$ where $F(X)=f(x).g(x)$ needs to be $F'(x) = \lim\limits_{h \to 0} \frac {f(x+h)-f(x)}{h}. \lim\limits_{h \to 0} \frac {g(x+h)-g(x)}{h}$ i.e. $f'(x).g'(x)$ but why it is $F'(x) = f(x).g'(x)+g(x).f'(x)$

Apologies if it sounds like a stupid question

Regards, Siddartha C.S

Answer 1

Definition: $$f'(x)=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$$

Let $p(x)=f(x)\cdot g(x)$, then:

$$p'(x)=\lim_{h\to 0}\frac{p(x+h)-p(x)}{h}=$$ $$=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=$$ $$=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}=$$ $$=\lim_{h\to 0}\frac{[f(x+h)-f(x)]g(x+h)+f(x)[g(x+h)-g(x)]}{h}=$$ $$=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\lim_{h\to 0}g(x+h)+\lim_{h\to 0}f(x)\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=$$ $$=f'(x)g(x)+f(x)g'(x)$$

Answer 2

Try with linear (affine) functions first to see why.

That is, let $f(x+h) = f(x)+f' h$, $g(x+h) = g(x)+g' h$, then $f(x+h)g(x+h)-f(x)g(x) = (f' g(x)+ f(x)g')h + f'g'h^2$