3.6 - The Chain Rule
And now, the chain rule! The chain rule comes into play whenever we want to take the derivative
of a composition of functions. In other words, how would we go about calculating [f(g(x))]?
First, let’s replace g(x) with u. It’s a little nicer to see this way. So, we want to eventually
calculate dx [f(u)]. Well, this isn’t just f (u), because u is also a function of x, and we can’t just
go ahead and do something like du[f(u)], because we want the derivative with respect to x.
Hmm...but what if we combine the two? So, we also have another derivative thrown in there, .
Now think of the \du" as a variable (I know it’s not, but just pretend for a second), and if we
multiply the two together, we can cancel out the du’s. In other words,
dy dy du
dx du dx
Bringing this all back to compositions of functions, we get:
[f(g(x))] = f (g(x))g (x)
Or, in other words, it’s the derivative of the outside function with the inside function plugged in
times the derivative of the inside function.
Example: f(x) = (x + 1) . Find f (x).0
f (x) = 5(x + 1) ▯ (2x)
The 5(x + 1) comes from the derivative of the outside function with the inside one plugged in,
and then the 2x is the derivative of what’s inside.
Example: f(x) = sin(x + 2x ▯ 5)2
0 2 3 2
f (x) = (3x + 4x)cos(x + 2x ▯ 5)
Guess what! You can derive the Quotient Rule a di▯erent way by using the product rule and
chain rule! Well, now that we’ve been introduced to the chain rule, let’s check it out!
d f(x) d ▯1
dx g(x) dx
= f (x)[g(x)] ▯1 + f(x)(▯1)[g(x)] ▯2g (x)
f (x)g(x) f(x)g(x)
= 2 ▯ 2
g (x) g (x)
gf ▯ fg 0
1 A much nicer way to ▯gure out the quotient rule rather than using the de▯nition of the derivative,
and by all means, if you can’t remember the quotient rule, you can always come back to
something like this!
Let’s do a few more examples.
Example: f(x) = (sin(x)cos(x)) . This problem would suck if you had to do product rule (three
f (x) = 2(sin(x)cos(x))[cos(x)cos(x) + sin(x)(▯sin(x))] = 2sin(x)cos (x) ▯ 2cos(x)sin (x) 3
Much nicer than doing the product rule 3 times!
What do you think might happen if we have the composition of the composition of functions? Ie,
Just work with it one function at a time. Chain rule says derivative of the outside with the inside
plugged in, times the derivative of the inside. So,
f(g(h(x))) = f (g(h(x)))g (h(x))h (x) 0
Example: f(x) = sin(cos(x )) 2