Product rule intuition

Question

Can anyone think of an intuitive explanation of the product rule? I'm not asking for a derivation. To me it seems like quite an un-untuitive result, as apposed to the chain-rule (which is ironically harder to derive).

Answer 1

If you think of $fg$ as giving the area of a rectangle with side lengths $f$ and $g$ (assuming $f,g$ are non-negative, otherwise just flip their signs), then if you change $x$ slightly then one side of the rectangle changes by an amount proportional to $df/dx$ and the other side of the rectangle changes by an amount proportional to $dg/dx$. So you get two little strips of changed area, which have total area proportional to $f'g + g'f$. The only question is whether the overlap of the two little strips matters, and it turns out that it doesn't because the overlap of the two strips has infinitesimal area which is a product of two infinitesimals (really really small), whereas the two strips have infinitesimal area which is a product of only one infinitesimal each (really small, but much larger than the overlap).

Answer 2

I like to think of it in terms of units and symmetry. We know that, unit-wise, the false formula

$$\frac{d}{dt} (f g) = \frac{df}{dt} \frac{dg}{dt}$$

cannot hold, because, if say $f$ and $g$ each represent distance and $t$ time, the LHS has units of distance^2/time, while the RHS has units distance^2/time^2.

Thus, the product rule for derivatives, in order to keep units, must be of the form $f dg/dt$ or vice-versa. But because of the symmetry of the original form, the product rule should also be symmetric, i.e. $f dg/dt + g df/dt$. As there is no other symmetric form that keeps the units, this should be the product rule, at least intuitively.

Answer 3

$$(x+\delta x)(y+\delta y)=xy + (x\delta y+y\delta x) + O(\delta^2)$$

Answer 4

If you know that $(\ln x)' = \dfrac1{x}$ and the chain rule, then $(\ln (fg))' = \dfrac{(fg)'}{fg} $ and $(\ln(fg))' =(\ln f + \ln g)' =\dfrac{f'}{f}+\dfrac{g'}{g} $.

Equating these two, $\dfrac{(fg)'}{fg} =\dfrac{f'}{f}+\dfrac{g'}{g} $ so that $(fg)' =f'g+g'f $.

The nice thing about this is that the quotient rule follows just as easily (with just copy and paste):

$(\ln (f/g))' = \dfrac{(f/g)'}{f/g} $ and $(\ln(f/g))' =(\ln f - \ln g)' =\dfrac{f'}{f}-\dfrac{g'}{g} $.

Equating these two, $\dfrac{(f/g)'}{f/g} =\dfrac{f'}{f}-\dfrac{g'}{g} $ so that $(f/g)' =(f/g)(\dfrac{f'}{f}-\dfrac{g'}{g}) =\dfrac{f'}{g}-\dfrac{fg'}{g^2} =\dfrac{f'g-fg'}{g^2} $.

And, of course, this allows the expression of the derivative of any number of products and quotients:

If $F(x) =\prod_{k=1}^n f_i^{a_i}(x) $, then $\ln(F(x)) =\sum_{k=1}^n a_i\ln f_i(x) $ so $\ln(F(x))' =\sum_{k=1}^n a_i(\ln f_i(x))' =\sum_{k=1}^n a_i\dfrac{f_i'(x)}{f_i(x)} $ and $\ln(F(x))' =\dfrac{F'(x)}{F(x)} $ so that $F'(x) = F(x)\sum_{k=1}^n a_i\dfrac{f_i'(x)}{f_i(x)} = \sum_{k=1}^n a_iF_i(x)f_i'(x) $ where $F_i(x) =\dfrac{F(x)}{f_i(x)} =f_i^{a_i-1}(x)\prod_{k=1, k\ne i}^n f_i^{a_i}(x) $.

Answer 5

For me intuition for product rule, as well as a couple of other techniques, comes from multi-variable calculus. Specifically for the product rule, take a function of two variables $f(y,z)$ and consider the formula for the differential of $f$:

$df=\partial f/\partial y \cdot dy + \partial f/\partial z \cdot dz$

The above is just a generalization of the chain rule, and IMO is very intuitive.

Now suppose that $f$ splits into a product of two functions, each being a function of just one of the variables: $f=g(y)\cdot h(z)$. Because $g(y)$ is constant with respect to $z$ and $h(z)$ is constant with respect to $y$ and differentiation is linear we have:

$d(g\cdot h)(y,z) = h(z)\cdot g'(y) \cdot dy + g(y)\cdot h'(z) \cdot dz$

Finally, it remains to consider what happens when both $y$ and $z$ have the same value $x$:

$d(g\cdot h)(x,x) = h\cdot g'(x) \cdot dx + g\cdot h'(x) \cdot dx$

and divide both sides by $dx$:

$(g\cdot h)'=hg'+gh'$

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Different people have different notions about what is intuitive. For me a part of being intuitive is the ability of immediately detect pattern and use it in other circumstances, and this approach goes well beyond the product rule. To understand why the above technique is useful try to compute the derivative of functions such as $f(x)=x^x$. This is very easy to do by replacing different occurrences of $x$ with separate variables, computing the partial derivatives, adding them up and setting all the variables to the same value $x$.

Answer 6

A very fundamental intuition mentioned to highschoolers inorder to make the product rule feel more natural for them (only for evaluation purpose, unrelated to actual calculus):

Differentiate one function, hold the rest constant. Rinse and repeat for each term.

$$(uv)' = \underbrace{uv'}_{u \text{ held constant}} + \underbrace{vu'}_{v\text{ held constant}}$$

Similarly, $$(abc)' = a'bc + ab'c + abc'\\ (abcd)' = a'bcd + ab'cd + abc'd + abcd'\\ \dots\\ (ab\dots yz)' = (a'b\dots yz) + (ab'\dots yz) + \dots + (ab\dots y'z) + (ab \dots yz') $$ Generally, we can say for functions $u_k(x)$, $k \in \mathbb N$, $$ \left(\prod_{i=1}^{n} u_i\right)' = \sum_{j =1}^{n} \left(\left ( \prod_{i=1}^n u_i \right)\cdot \frac{u_j'}{u_j}\right)$$ This idea helps many students fearlessly apply the product rule. Also, with this idea in mind remembering the Quotient Rule can be avoided for it is merely a corollary of the product rule.