Why is $\lim_{x\to a}{[f(x)\cdot g(x)]}=\lim_{x\to a}{f(x)}\cdot\lim_{x\to a}{g(x)}$, but $(fg)' = f'g+fg'$

Question

For example the limit of 2 functions multiplied together is a straightforward equivalent of the multiplication of each function separately limited: $$\lim_{x\to a}{[f(x)\cdot g(x)]}=\lim_{x\to a}{f(x)}\cdot\lim_{x\to a}{g(x)}$$ but not so with derivatives. The product rule requires: $$(fg)'=f'g + fg'$$ How can this be?

Answer 1

The derivative of $fg$ is not a product of limits of $f$ and $g$. Recall that $$f'(x) = \lim_{y \to x} \frac{f(y) - f(x)}{y-x}$$ So using our result for the limit of a product would give us \begin{align*} f'(x)g'(x) =& \bigg[\lim_{y \to x} \frac{f(y) - f(x)}{y-x}\bigg] \cdot \bigg[\lim_{y \to x} \frac{g(y) - g(x)}{y-x} \bigg]\\ \neq& \lim_{y \to x} \frac{f(y)g(y) - f(x)g(x)}{x-y} = (fg)'(x) \end{align*} in general.

Answer 2

These are limits of different things, and there is no reason to expect $(fg)'$ to be $f'g'$. In fact, there is, a priori, not even a reason to expect the first relation that you wrote down to be true, it requires rigorous proof (which, if you are learning this in school, you were probably not given).

Does the following example help you? The limit $\lim_{x\to a}\left((f(x)-5)(g(x)-5)\right)$ is certainly not equal to $\lim_{x\to a}(f(x)g(x)-5)$, so you cannot always expect to be able to pull products into limits in a simple minded way.

Answer 3

Yes a derivative is a limit. And, yes, $\lim_{x\to a} f(x)g(x) = \lim_{x\to a} f(x)\lim_{x\to a}g(x)$

But the derivative is something a little bit more complicated than what you suggest.

$\frac {d}{dx} f(x) = f'(x) = \lim_\limits{h\to 0} \frac {f(x+h) - f(x)}{h}$

$\frac {d}{dx} f(x)g(x) = $$\lim_\limits{h\to 0} \frac {f(x+h)g(x+h) - f(x)g(x)}{h}\\ \lim_\limits{h\to 0} \frac {f(x+h)g(x+h) -f(x+h)g(x) +f(x+h)g(x) - f(x)g(x)}{h}\\ \lim_\limits{h\to 0} \frac {f(x+h)(g(x+h) -g(x)) +(f(x+h) - f(x))g(x)}{h}\\ \lim_\limits{h\to 0} f(x+h)\frac {g(x+h) -g(x)}{h} + \lim_\limits{h\to 0}\frac {f(x+h) - f(x)}{h}g(x)\\ f(x) g'(x) + f'(x)g(x)$

Answer 4

It's because of how we multiply fractions.

The derivative of $f(x)$ at a point $x=a$ is $$f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ The derivative of $g(x)$ at the same point is $$g'(a) = \lim_{x\to a}\frac{g(x)-g(a)}{x-a}$$

If you multiplied the first two expressions together you'd get $$f'(a)g'(a) = \lim_{x \to a} \frac{\left(f(x)-f(a)\right)\left(g(x)-g(a) \right)}{(x-a)^2}=\lim_{x\to a}\frac{f(x)g(x)-f(a)g(x)-g(a)f(x)+f(a)g(a)}{(x-a)^2}$$

However, this is not the derivative of $f(x)g(x)$ at $x=a$. That would be given by

$$(fg)'(a) = \lim_{x\to a}\frac{f(x)g(x)-f(a)g(a)}{x-a}$$

You can see that the expression for $f'(a)g'(a)$ has extra unwanted terms in the numerator, the wrong sign on the last term in the numerator, and the wrong power of $(x-a)$ in the denominator.

Answer 5

Here is a simple model just based on products, which may help you to understand. Note that $$(x+p)(y+q)=xy+xq+py+pq$$ so that $$(x+p)(y+q)-xy=xq+py+pq$$

Here we can see how the change in the product (and therefore the derivative of the product) is related to the cross terms in the product - if $p$ and $q$ are both small, their product will be insignificant.