Why The Chain Rule Works

For the past 3 years, some of my A students in Calculus have returned as Teaching Assistants for later classes.  I’ve learned a lot from them and their questions, and they do an excellent job helping new students understand.  Recently, one of my TAs pointed out to me that while my calculus class had a great visual explanation of the Product Rule, it lacked a visual explanation of the Chain Rule… and then he offered a solution he’d found.  It’s brilliant!  Grab a pencil and paper, and follow the wisdom of PhysicsForum’s epenguin:

[f(g(x)]’ = f'[g(x)]g'(x)
On a piece of paper, draw rectangular axes. Right top quadrant is going to represent g (example of positive x and positive g(x) ). Sketch a smooth function in that quadrant. Horizontal axis is x, vertical is g(x).

Now turn paper 90° clockwise. The g axis is still the g axis but now it is horizontal. Draw another smooth curve in the now top quadrant – that represents f. The now vertical axis (think of it as a different piece of paper) is the f of the corresponding points of the now horizontal axis. I.e. is f(g).

So, turning the paper back to its original orientation, starting from a point x, if you draw a vertical line to the first curve, and from the meeting point a horizontal line to the second curve, you get to f[g(x)] on the second curve (turning the paper again). Take a point just above x and do the same thing again. You get a thin strip of width dx (think as) and then another thin strip of width g’(x).dx and rotating the paper again you should be able to see that it corresponds to an increase of f (i.e. df or in full d[f(g(x))]) of [f(g(x)]’ = f'[g(x)]g'(x)dx


Does your drawing look like this one?


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s