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Mathematics

MAT237Y1

Dan Dolderman

Fall

Description

Calc 1
Tues 9/30/08
3.6 - The Chain Rule
And now, the chain rule! The chain rule comes into play whenever we want to take the derivative
d
of a composition of functions. In other words, how would we go about calculating [f(g(x))]?
dx
First, let’s replace g(x) with u. It’s a little nicer to see this way. So, we want to eventually
d 0
calculate dx [f(u)]. Well, this isn’t just f (u), because u is also a function of x, and we can’t just
d
go ahead and do something like du[f(u)], because we want the derivative with respect to x.
du
Hmm...but what if we combine the two? So, we also have another derivative thrown in there, .
dx
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:
Chain Rule:
d
[f(g(x))] = f (g(x))g (x)
dx
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
Solution:
f (x) = 5(x + 1) ▯ (2x)
2 4
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
Solution:
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
= f(x)[g(x)]
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
=
g2
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
times!!!).
Solution:
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,
what’s
d
f(g(h(x)))?
dx
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,
d
f(g(h(x))) = f (g(h(x)))g (h(x))h (x) 0
dx
Example: f(x) = sin(cos(x )) 2
Solution:

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