Product Rule

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We develop the technique for decomposing the product of functions for derivatives, introducing the “Product Rule”.

Video Lecture

Text and Additional Details

We saw earlier that the sum and difference rules for derivatives were (eventually) simply an application of the sum and difference rules for limits. Indeed, even the constant multiple rule for derivatives was simply an application of the constant multiple rule for limits.

Unfortunately however, taking a derivative of a product of functions is not as simple. Indeed, to come up with an elegant (... ish) rule for this, we need to use the technique of “adding zero ... cleverly.”

As usual, we start by considering the computation we want to understand. In particular, we start by looking at the limit of the difference quotient for $f(x)g(x)$ .

$\lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x}$

Considering the corresponding limit laws, we would want something like the derivative of $f(x)g(x)$ to be $f'(x)g'(x)$ ... unfortunately that’s just not true. Indeed, even a basic example demonstrates this. Consider for a moment, the following example: let $f(x) = x$ and $g(x) = x$ . On the one hand, we know from the polynomial derivative formula that $f'(x)=1x^{1-1} = 1$ and similarly $g'(x) = 1$ . On the other hand, $f(x)g(x) = x\cdot x = x^2$ and by the same polynomial derivative formula, the derivative of $x^2$ is $(2)x^{2-1} = 2x$ . Thus the derivative of $f(x)g(x)$ would be the derivative of $x^2$ which is $2x$ , but $f'(x)g'(x) = 1\cdot 1 = 1 \neq 2x$ ... so this (clearly) doesn’t work and thus we aren’t so lucky.

So, with our dream of an easy rule shattered, we are back to being stuck with this difference quotient... so what do we do? Here comes the rabbit out of a hat moment; we will add and subtract the expression $f(x + \Delta x)g(x)$ to the top of our fraction. Why this particular expression? The unsatisfying answer would be “because it works” but really, the answer is much more nuanced than that. To see why this choice of term works, we need to observe that we currently have two very unrelated parts in the numerator; an expression entirely in terms of $x$ , (i.e. $f(x)g(x)$ ), and an expression entirely in terms of $x + \Delta x$ , (i.e. $f(x + \Delta x)g(x + \Delta x)$ ). This new expression we are introducing allows us to bridge the gap between these two expressions, with a surprising result!

$\displaystyle \lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x}$
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x + \Delta x) \left ( - f(x + \Delta x)g(x) + f(x + \Delta x)g(x) \right )- f(x)g(x)}{\Delta x}$	Added Zero “Cleverly”.
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0} \frac {\bigg (f(x + \Delta x)g(x + \Delta x) - f(x + \Delta x)g(x)\bigg ) + \bigg (f(x + \Delta x)g(x) - f(x)g(x)\bigg )}{\Delta x}$	Regrouped terms.
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)\bigg (g(x + \Delta x) - g(x)\bigg ) + g(x)\bigg (f(x + \Delta x - f(x)\bigg )}{\Delta x}$	Factor out common terms.
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)\bigg (g(x + \Delta x) - g(x)\bigg )}{\Delta x} + \lim \limits _{\Delta x\rightarrow 0}\frac {g(x)\bigg (f(x + \Delta x - f(x)\bigg )}{\Delta x}$	Split limit over sum.
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0}f(x + \Delta x) \cdot \lim \limits _{\Delta x\rightarrow 0} \frac {\bigg (g(x + \Delta x) - g(x)\bigg )}{\Delta x} + g(x)\lim \limits _{\Delta x\rightarrow 0}\frac {\bigg (f(x + \Delta x - f(x)\bigg )}{\Delta x}$	Split limit over product.
	Factored out constant multiple $g(x)$ .
$\displaystyle = f(x) \cdot \lim \limits _{\Delta x\rightarrow 0} \frac {\bigg (g(x + \Delta x) - g(x)\bigg )}{\Delta x} + g(x)\lim \limits _{\Delta x\rightarrow 0}\frac {\bigg (f(x + \Delta x - f(x)\bigg )}{\Delta x}$	Evaluated $\lim \limits _{\Delta x\rightarrow 0}f(x + \Delta x)$
$\displaystyle = f(x)g'(x) + g(x)f'(x)$	Turn Dif. Quot. into Prime notation.

So, it turns out that adding and subtracting that (admittedly strange) bridging expression allowed us to do some clever factoring and break apart the difference quotient using limit laws to ultimately record the following derivative law:

Product Rule Let $f(x)$ and $g(x)$ be differentiable functions. Then: $\frac {d}{dx} \left [ f(x)g(x) \right ] = f'(x)g(x) + f(x)g'(x)$

Notice that, since addition and multiplication are commutative, it doesn’t matter what order we write any of this law in. As long as we make sure that each term has exactly one factor that is the original function and the other factor is the derivative of the other function, we’re good!

We have seen that, although the derivative law for products isn’t quite as nice as we would like, it isn’t too bad after all. With a bit of adding zero cleverly, followed by some careful factoring and manipulation, we arrived at our derivative product rule; the derivative of the product $f(x)g(x)$ is the sum of $f'(x)g(x)$ and $g'(x)f(x)$ . The fact that this rule is stated using only sums and products is deceptively nice, as it means we don’t have to worry about remembering the correct order; we can rearrange it however we like as long as we make sure each term has one derivative, and one original function. This is noteworthy as the quotient rule isn’t quite so nice.