# Is there an intuitive way to think about the product rule for derivatives?

$$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$$

Is there a reason why this makes sense? I mean really, if you break things down in terms of slopes of tangent lines and what not, there is a lot going on here. Is there some geometric intuition going on behind this that I've been blind to all of these years?

• There's a geometric argument using the area of a rectangle $A(x)=u(x) v(x)$. Here's a post of that math.stackexchange.com/questions/397554/…
– Joe
Sep 16 at 22:20
• I found this very helpful, it's one of my favourite parts of the series Sep 16 at 22:22
• I can't find a good link to illustrate this, but my intuition about this has always been based on thinking about a drive chain with two elliptical chain wheels. The gain of each chain wheel contributes a gain that is proportional to its gain at that position and the position of the other one. Sep 16 at 22:29
• @RobArthan My feeble mind is having problems seeing what you are explaining Sep 16 at 22:30
• I am trying to explain the product rule for derivatives as per your question. Do you ride a bicycle? Draw some pictures of what it would be like if the chain wheels were elliptical. Think about how your work on the pedals would affect the position of the road wheel. Sep 16 at 22:35

Suppose you have a rectangle with dimensions $$f$$ by $$g$$ units. Then the area $$A = fg$$. Now suppose $$f$$ is increased by $$df$$ and $$g$$ is increased by $$df$$. The new rectangle has area

$$(f+df)(g+dg) = fg +f(dg) + g(df) +(df)(dg).$$

So the area was increased by

$$f(dg) + g(df) +(df)(dg).$$

Since we're waving our hands here, we note that the last term is not just tiny, but tiny-squared, so we neglect it and note that the change in area is really close to $$f(dg) + g(df).$$

• Nice, grabbing this for my toolkit for the next time I get to teach calculus.
– Alan
Sep 16 at 22:52
• @Alan A graphic of the rectangle is almost a proof-without-words. There one on this page: web.mit.edu/wwmath/calculus/differentiation/products.html Sep 16 at 23:43