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Chain
Rule

We’ve been using the standard chain rule for functions of one variable throughout the last couple
of sections
...
Before we
actually do that let’s first review the notation for the chain rule for functions of one variable
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F ( x ) = f ( g ( x ))

F′ ( x ) = f ′ ( g ( x ) ) g′ ( x )

There is an alternate notation however that while probably not used much in Calculus I is more
convenient at this point because it will match up with the notation that we are going to be using in
this section
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If

y = f ( x)

and

x = g (t )

then

dy dy dx
=
dt dx dt

dy
really does make sense here since if we were to plug in for x then y
Notice that the derivative
dt
really would be a function of t
...

Okay, now that we’ve got that out of the way let’s move into the more complicated chain rules
that we are liable to run across in this course
...
So, let’s start this discussion off
with a function of two variables, z = f ( x, y )
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We will be looking at two distinct cases prior to generalizing the
whole idea out
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dt
This case is analogous to the standard chain rule from Calculus I that we looked at above
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The chain rule for this case is,

dz ∂f dx ∂f dy
=
+
dt ∂x dt ∂y dt

So, basically what we’re doing here is differentiating f with respect to each variable in it and then
multiplying each of these by the derivative of that variable with respect to t
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Let’s take a look at a couple of examples
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dt
(a) z = xe xy , x = t 2 , y = t −1 [Solution]

( )

(b) z = x 2 y 3 + y cos x , x = ln t 2 , y = sin ( 4t ) [Solution]
Solution
(a) z = xe xy , x = t 2 , y = t −1
There really isn’t all that much to do here other than using the formula
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However, we should probably go ahead and
substitute in for x and y as well at this point since we’ve already got t’s in the derivative
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For comparisons sake
let’s do that
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Note however, that often it will actually be more work to do the
substitution first
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dz
⎛2⎞
3
= ( 2 xy − y sin x ) ⎜ ⎟ + ( 3 x 2 y 2 + cos x ) ( 4 cos ( 4t ) )
dt
⎝t⎠
=

4sin 3 ( 4t ) ln t 2 − sin ( 4t ) sin ( ln t 2 )

t

(

+ 4 cos ( 4t ) 3sin ( 4t ) ⎡⎣ln t ⎤⎦ + cos ( ln t 2 )
2

2

2

)

Note that sometimes, because of the significant mess of the final answer, we will only simplify
the first step a little and leave the answer in terms of x, y, and t
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[Return to Problems]

Now, there is a special case that we should take a quick look at before moving on to the next case
...


dz
2
3
Example 2 Compute
for z = x ln ( xy ) + y , y = cos ( x + 1)
dx
Solution
We’ll just plug into the formula
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∂z
∂z
and

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Here is the chain rule for both of these cases
...
Here is a quick
example of this kind of chain rule
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∂s
∂t
Solution

∂z

...

Now the chain rule for
∂t
∂z
= 2e2 r sin ( 3θ ) ( s − 2t ) + 3e2 r cos ( 3θ )
∂t

(

s

)

(
3te

t
s2 + t 2

2 st −t 2

(

) cos 3 s 2 + t 2

)

s2 + t 2

Okay, now that we’ve seen a couple of cases for the chain rule let’s see the general version of the
chain rule
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Then for any variable ti , i = 1, 2,…, m we have the
following,

∂z ∂z ∂x1 ∂z ∂x2
=
+
+
∂ti ∂x1 ∂ti ∂x2 ∂ti

∂z ∂xn
+
∂xn ∂ti

Wow
...
There is actually an easier way to construct all the chain rules
that we’ve discussed in the section or will look at in later examples
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To see how these work let’s go back

∂z
and take a look at the chain rule for
given that z = f ( x, y ) , x = g ( s,t ) , y = h ( s,t )
...
For reference here is the chain rule for this case,

∂z ∂f ∂x ∂f ∂y
=
+
∂s ∂x ∂s ∂y ∂s

Here is the tree diagram for this case
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The first set of
branches is for the variables in the function
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We connect each letter
with a line and each line represents a partial derivative as shown
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To use this to get the chain rule we start at the bottom and for each branch that ends with the
variable we want to take the derivative with respect to (s in this case) we move up the tree until
we hit the top multiplying the derivatives that we see along that set of branches
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Note that we don’t usually put the derivatives in the tree
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Let’s write down some chain rules
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dw
(a)
for w = f ( x, y, z ) , x = g1 ( t ) , y = g 2 ( t ) , and z = g 3 ( t ) [Solution]
dt
∂w
(b)
for w = f ( x, y, z ) , x = g1 ( s,t, r ) , y = g 2 ( s,t, r ) , and z = g3 ( s,t, r )
∂r
[Solution]

Solution

dw
for w = f ( x, y, z ) , x = g1 ( t ) , y = g 2 ( t ) , and z = g3 ( t )
(a)
dt
So, we’ll first need the tree diagram so let’s get that
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[Return to Problems]

∂w
(b)
for w = f ( x, y, z ) , x = g1 ( s,t, r ) , y = g 2 ( s,t, r ) , and z = g3 ( s,t, r )
∂r
Here is the tree diagram for this situation
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We’ve now seen how to take first derivatives of these more complicated situations, but what
about higher order derivatives? How do we do those? It’s probably easiest to see how to deal
with these with an example
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2
∂θ
Solution
We will need the first derivative before we can even think about finding the second derivative so
let’s get that
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Here is the first derivative
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Since the two first order derivatives,
∂x

∂f
, are both functions of x and y which are in turn functions of r and θ both of these terms
and
∂y
are products
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These are both chain rule
⎜ ⎟ and
∂θ ⎝ ∂y ⎠
∂θ ⎝ ∂x ⎠
problems again since both of the derivatives are functions of x and y and we want to take the
derivative with respect to θ
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∂
∂
∂
( f ) = −r sin (θ ) ( f ) + r cos (θ ) ( f )
∂θ
∂x
∂y

(1)

Note that all we’ve done is change the notation for the derivative a little
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This however is exactly what we need to do the two new derivatives we need above
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To do this we’ll simply replace all the f ‘s in (1) with the first order partial derivative that we want
to differentiate
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∂ ⎛ ∂f ⎞
Here is the use of (1) to compute
⎜ ⎟
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⎝ ∂y ⎠

∂
∂θ

⎛ ∂f ⎞
∂ ⎛ ∂f ⎞
∂ ⎛ ∂f ⎞
⎜ ⎟ = −r sin (θ ) ⎜ ⎟ + r cos (θ ) ⎜ ⎟
∂x ⎝ ∂y ⎠
∂y ⎝ ∂y ⎠
⎝ ∂y ⎠
∂2 f
∂2 f
= −r sin (θ )
+ r cos (θ ) 2
∂x∂y
∂y

The final step is to plug these back into the second derivative and do some simplifying
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The final topic in this section is a revisiting of implicit differentiation
...
Let’s start out with
the implicit differentiation that we saw in a Calculus I course
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In a Calculus I

dy
and this was often a fairly messy process
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We’ll start
by differentiating both sides with respect to x
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Here are the results of that
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Note as well that in order to simplify the formula we switched back to
using the subscript notation for the derivatives
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dy
Example 6 Find
for x cos ( 3y ) + x 3 y 5 = 3x − e xy
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x cos ( 3y ) + x 3 y 5 − 3x + e xy = 0

Now, the function on the left is F ( x, y ) in our formula so all we need to do is use the formula to
find the derivative
...
It would have taken much longer to do this using the old Calculus I way of doing
this
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In
these cases we will start off with a function in the form F ( x, y, z ) = 0 and assume that

∂z
∂z
z = f ( x, y ) and we want to find
and/or

...
We will differentiate both sides with respect to x and we’ll need
∂x
to remember that we’re going to be treating y as a constant
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Here is this derivative
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The
second is because we are treating the y as a constant and so it will differentiate to zero
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Let’s take a quick look at an example of this
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∂y
∂x
Solution
This was one of the functions that we used the old implicit differentiation on back in the Partial
Derivatives section
...

First let’s get everything on one side
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2 x sin ( 2 y − 5 z ) + 6 yz sin ( 6 zx )
∂z
=−
∂x
−5 x 2 cos ( 2 y − 5 z ) + 6 yx sin ( 6 zx )

2 x 2 cos ( 2 y − 5 z ) − cos ( 6 zx )
∂z
=−
∂y
−5 x 2 cos ( 2 y − 5 z ) + 6 yx sin ( 6 zx )
If you go back and compare these answers to those that we found the first time around you will
notice that they might appear to be different
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