The Chain Rule Explained
It is common sense to take a method and try it. If it fails, admit it frankly and try another. But above all, try something.
-Franklin D. Roosevelt, 32nd United States President
We all know how to take a derivative of a basic function (such as y x 2 2 x 8
or y ln x ), right? Now, what do you do if you are not given a function that
straightforward? What would you do if the function was: y ln x 2 2 x 8 ? How would
you find the derivative of that? In case you haven‟t guessed it yet, it would be the Chain
Rule!
Show Me How
dy
dx
dy du
du dx
(1)
What does that above equation mean? That‟s the Chain Rule in “math-speak”.
dy
, is the basic form of derivate and is read as “the derivative of „y‟ with
The first term,
dx
respect to „x‟.” As a reminder, that means you have a function, y (x), that is in terms of
“x”. When you take the derivative of that function you only care about the “x” terms.
dy
dy
, is just a variation of
That‟s it. The second term,
and is read, almost, the same
du
dx
du
, is “the derivative of
way: “the derivative of „y‟ with respect to „u‟.” The last term,
dx
„u‟ with respect to „x‟” (now, don‟t let the variable change scare you—our function can
be called anything you want, we are simply using “u” in this case).
dy
We separate
into those two other terms to make the Chain Rule a bit easier to
dx
figure out. Let‟s go back to the example I first gave you: y ln x2 2 x 8 . We know
the components—ln x and x 2 2 x 8 —are easy to take the derivatives of, right? So,
let‟s break down ln x 2 2 x 8 into those components, but make one slight change:
u x x2 2x 8
So, replacing „u‟ with x 2 2 x 8 you get a new function:
y u ln u
See the variable change? All I did was call all that junk in the parenthesis of
ln x 2 2 x 8 “u” then I just made a note saying “hey, my function „u‟ is done with
respect to „x‟ and my function „y‟ is done with respect to the variable „u‟.” If I were to
plug u(x) in for “u” in my y(u) function I should get back the original equation, right?
yux
ln x 2 2 x 8
Another way of writing that is:
y  u ln x2 2 x 8
Does that seem familiar? Looks a lot like College Algebra and the Composition section
(think: f  g or “fog”). Why am I bringing up College Algebra for a Calculus problem?
What does College Algebra have to do with Calculus? EVERYTHING! Math is a subject
that builds from previous lessons, you simply cannot do exponents without knowing how
to multiply, you cannot subtract without knowing how to add, and you cannot fully
understand the Chain Rule without remembering Compositions!
Ok, so back to the problem at hand: the derivative of ln x 2 2 x 8 . To recap,
I‟ve given you the following:
dy dy du
dx du dx
y x ln x 2 2 x 8
u x x2 2x 8
y u ln u
And remember that y x comes from plugging “u” in for “x” for the original equation.
dy
du
) and u x (also known as
)
Now, I can find the derivative of y u (also known as
du
dx
really easily.
dy 1
du u
du
2x 2 2 x 1
dx
dy
And, from equation (1), I can find the derivative ( ) of y ln x2 2 x 8 easily.
dx
dy dy du
dx du dx
dy 1
2x 1
dx u
Plug u x x2 2 x 8 back into the equation, replacing “u” with x 2 2 x 8 and place
2 x 1 into the numerator (due to multiplication) and it will give us the derivative of
y ln x2 2 x 8 :
dy
2( x 1)
2
dx x 2 x 8
Examples
Example 1
If y
3
x 2 1 , find y
Solution
Using the Chain Rule we will get the following:
dy
dx
x2 1
u ( x)
y (u ) u 3
du
2x
dx
dy
3u 2
du
dy
y
dx
dy
y
dx
dy
y
dx
dy du
du dx
3u 2 2 x
6x x2 1
2
Example 2
e3 x , find y
If y
dy
dx
Solution
Using the Chain Rule, we get the following:
u ( x) 3 x
eu
y (u )
du
dx
dy
du
3
eu
dy
dx
dy
dx
dy
dx
y
y
y
dy du
du dx
eu 3
3e3 x
Example 3
If y
ln x 4
2
2 x 2 3 , find y
dy
dx
Solution
We will still be using the chain rule to figure out this problem, but I will also be using a
rule of logarithms (you remember those?) to make solving this a bit easier:
u x x4 2x2 3
yu
du
dx
ln u 2
4 x3
2 ln u
4x
4 x( x 2 1)
dy
du
y
y
2
u
dy dy du
dx du dx
dy 2
8x x 2 1
4x x 2 1
dx u
x 4 2x 2 3
Additional Information
2
1
u
1. http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html (Visual
Calculus - Chain Rule)
2. http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/
ChainRule.html (Differentiation Using the Chain Rule)
3. http://www.math.hmc.edu/calculus/tutorials/chainrule/ (HMC Mathematics
Online Tutorial - The Chain Rule)