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Partial
Derivatives

Now that we have the brief discussion on limits out of the way we can proceed into taking
derivatives of functions of more than one variable
...

Recall that given a function of one variable, f ( x ) , the derivative, f ′ ( x ) , represents the rate of
change of the function as x changes
...
The problem with functions
of more than one variable is that there is more than one variable
...
For instance, one variable could be changing
faster than the other variable(s) in the function
...

We will need to develop ways, and notations, for dealing with all of these cases
...
We will deal with allowing multiple variables to change in a
later section
...
Let’s start off this discussion with a fairly simple function
...

We’ll start by looking at the case of holding y fixed and allowing x to vary
...
Doing this will give us a function involving only x’s and
we can define a new function as follows,

g ( x ) = f ( x,b ) = 2x b

2 3

Now, this is a function of a single variable and at this point all that we are asking is to determine
the rate of change of g ( x ) at x = a
...
Here is the rate of change of
the function at ( a,b ) if we hold y fixed and allow x to vary
...
We will now hold x fixed and allow y to vary
...
Since we are holding x fixed it must be fixed at x = a and so we can define a new
function of y and then differentiate this as we’ve always done with functions of one variable
...

Just as with functions of one variable we can have derivatives of all orders
...

Note that the notation for partial derivatives is different than that for derivatives of functions of a
single variable
...
However, with partial derivatives we will always need to remember the variable that we
are differentiating with respect to and so we will subscript the variable that we differentiated with
respect to
...

Note as well that we usually don’t use the ( a,b ) notation for partial derivatives
...
So, the partial derivatives from above will
more commonly be written as,

f x ( x, y ) = 4xy

3

and

f y ( x, y ) = 6x y
2

2

Now, as this quick example has shown taking derivatives of functions of more than one variable
is done in pretty much the same manner as taking derivatives of a single variable
...
Likewise, to compute f y ( x, y ) we will treat all the x’s as constants

and then differentiate the y’s as we are used to doing
...

Since we can think of the two partial derivatives above as derivatives of single variable functions
it shouldn’t be too surprising that the definition of each is very similar to the definition of the
derivative for single variable functions
...


f ( x + h, y ) − f ( x, y )
f x ( x, y ) = lim
h →0
h

f ( x, y + h ) − f ( x, y )
f y ( x, y ) = lim
h→0
h

Now let’s take a quick look at some of the possible alternate notations for partial derivatives
...


f ( x)

⇒

f ( x, y )

⇒

df
f ′( x) =
dx
∂f
∂f
fx ( x) =
& fy ( x) =
∂x
∂y

Okay, now let’s work some examples
...

This is important because we are going to treat all other variables as constants and then proceed
with the derivative as if it was a function of a single variable
...


Example 1 Find all of the first order partial derivatives for the following functions
...
The partial derivative with respect to x is,

f x ( x, y ) = 4x 3

Notice that the second and the third term differentiate to zero in this case
...
It’s a constant and we know that constants always
differentiate to zero
...

Remember that since we are differentiating with respect to x here we are going to treat all y’s as
constants
...

Now, let’s take the derivative with respect to y
...
Here is
the partial derivative with respect to y
...
Let’s do the partial
derivative with respect to x first
...
This means that the second and fourth terms will differentiate to zero
since they only involve y’s and z’s
...

Here is the partial derivative with respect to x
...
In this case all x’s and z’s will be treated as constants
...
Here is the
derivative with respect to y
...
Since only one of the terms involve z’s this will
be the only non-zero term in the derivative
...
Here is the derivative with respect to z
...
Before taking the derivative let’s rewrite
the function a little to help us with the differentiation process
...
It will work the same way
...


∂h 7 ⎛ 2 s ⎞ 4 − 73 2t 7 4 − 73
hs ( s, t ) =
=t ⎜ 2 ⎟− s =
− s
∂s
s 7
⎝s ⎠ 7
∂h
ht ( s, t ) =
= 7t 6 ln ( s 2 ) − 27t −4
∂t
Remember how to differentiate natural logarithms
...
The product rule will work the same way
here as it does with functions of one variable
...

Let’s start out by differentiating with respect to x
...
Here is the derivative with respect to x
...
We will be looking at the chain rule for
some more complicated expressions for multivariable functions in a latter section
...

Also, don’t forget how to differentiate exponential functions,

( )

d f ( x)
f ( x)
′
e
= f ( x)e
dx
Now, let’s differentiate with respect to y
...
Therefore, since x’s are considered
to be constants for this derivative, the cosine in the front will also be thought of as a
multiplicative constant
...


⎛ 4 ⎞ x 2 y −5 y 3
f y ( x, y ) = ( x − 15 y ) cos ⎜ ⎟ e
⎝ x⎠
2

2

[Return to Problems]

Example 2 Find all of the first order partial derivatives for the following functions
...
Since there isn’t too much to this one, we will
simply give the derivatives
...
In this case we do have a quotient, however, since the x’s and y’s only appear in the
numerator and the z’s only appear in the denominator this really isn’t a quotient rule problem
...
In both these cases the z’s are constants and
so the denominator in this is a constant and so we don’t really need to worry too much about it
...


sin ( y )
g x ( x, y , z ) =
z2

x cos ( y )
g y ( x, y , z ) =
z2

Now, in the case of differentiation with respect to z we can avoid the quotient rule with a quick
rewrite of the function
...


g ( x, y, z ) = x sin ( y ) z −2

2 x sin ( y )
g z ( x, y, z ) = −2 x sin ( y ) z = −
z3
−3

We went ahead and put the derivative back into the “original” form just so we could say that we
did
...

[Return to Problems]

(c) z =

x 2 + ln ( 5x − 3y 2 )

In this last part we are just going to do a somewhat messy chain rule problem
...

Here are the two derivatives,

(

)

(

)

1 2
z x = x + ln ( 5 x − 3 y 2 )
2
1 2
= x + ln ( 5 x − 3 y 2 )
2

∂ 2
x + ln ( 5 x − 3 y 2 )
∂x
1
− ⎛
⎞
5
2
⎜ 2x +
2 ⎟
5x − 3 y ⎠
⎝
−

1
2

(

1
⎛
⎞
−
5
2
2
2
⎟
x
x
y
=⎜x+
+
ln
5
−
3
(
)
2
⎜
⎟
x
y
2
5
−
3
(
)
⎝
⎠
1
−
1 2
2
2 ∂
z y = x + ln ( 5 x − 3 y )
x 2 + ln ( 5 x − 3 y 2 )
2
∂y

)

(

(

)

(

)

1 2
= x + ln ( 5 x − 3 y 2 )
2

(

)

(

−

1
2

)

⎛ −6 y ⎞
⎜
2 ⎟
x
y
5
−
3
⎝
⎠

3y
2
2
x
x
y
=−
+
ln
5
−
3
(
)
2
5x − 3 y

)

−

1
2

[Return to Problems]

So, there are some examples of partial derivatives
...
So, if you can do Calculus I derivative you shouldn’t
have too much difficulty in doing basic partial derivatives
...
Before getting into implicit differentiation for multiple variable functions let’s
first remember how implicit differentiation works for functions of one variable
...

dx
Solution
Remember that the key to this is to always think of y as a function of x, or y = y ( x ) and so
whenever we differentiate a term involving y’s with respect to x we will really need to use the

dy
chain rule which will mean that we will add on a
to that term
...


dy
12 y
+ 7 x6 = 5
dx
3

dy
The final step is to solve for

...
If we have a function in terms of three variables x, y, and z we
will assume that z is in fact a function of x and y
...
Then whenever

∂z

...

∂y
Let’s take a quick look at a couple of implicit differentiation problems
...

∂x
∂y
3 2
5
2
3
(a) x z − 5xy z = x + y [Solution]
(b) x 2 sin ( 2 y − 5z ) = 1+ y cos ( 6zx ) [Solution]
Solution
(a) x3 z 2 − 5xy 5 z = x 2 + y 3

∂z

...

∂x
2 2
3 ∂z
5
5 ∂z
3x z + 2 x z − 5 y z − 5 xy
= 2x
∂x
∂x
Remember that since we are assuming z = z ( x, y ) then any product of x’s and z’s will be a
product and so will need the product rule!

∂z
Now, solve for

...


∂z
4
5 ∂z
= 3 y2
2 x z − 25 xy z − 5 xy
∂y
∂y
3
5 ∂z
2
4
x
z
xy
y
xy
z
−
=
+
2
5
3
25
(
) ∂y
3

∂z 3 y 2 + 25 xy 4 z
=
∂y
2 x 3 z − 5 xy 5
[Return to Problems]

(b) x 2 sin ( 2 y − 5z ) = 1+ y cos ( 6zx )

∂z

...
First let’s find
∂x
∂z ⎞
⎛ ∂z ⎞
⎛
2
2 x sin ( 2 y − 5 z ) + x cos ( 2 y − 5 z ) ⎜ −5 ⎟ = − y sin ( 6 zx ) ⎜ 6 z + 6 x ⎟
∂x ⎠
⎝ ∂x ⎠
⎝
Don’t forget to do the chain rule on each of the trig functions and when we are differentiating the

∂z

...
Now let’s solve for
∂x
∂z 2
∂z
2 x sin ( 2 y − 5 z ) − 5 x cos ( 2 y − 5 z ) = −6 zy sin ( 6 zx ) − 6 yx sin ( 6 zx )
∂x
∂x
∂z
2
2 x sin ( 2 y − 5 z ) + 6 zy sin ( 6 zx ) = ( 5 x cos ( 2 y − 5 z ) − 6 yx sin ( 6 zx ) )
∂x
2 x sin ( 2 y − 5 z ) + 6 zy sin ( 6 zx )
∂z
= 2
∂x 5 x cos ( 2 y − 5 z ) − 6 yx sin ( 6 zx )
∂z

...

Now let’s take care of
∂y
⎛
⎛ ∂z ⎞
∂z ⎞
x cos ( 2 y − 5 z ) ⎜ 2 − 5 ⎟ = cos ( 6 zx ) − y sin ( 6 zx ) ⎜ 6 x ⎟
∂y ⎠
⎝
⎝ ∂y ⎠
∂z
∂z
2
2
2 x cos ( 2 y − 5 z ) − 5 x cos ( 2 y − 5 z ) = cos ( 6 zx ) − 6 xy sin ( 6 zx )
∂y
∂y
∂z
2
2
−
−
=
−
xy
zx
x
y
z
zx
x
6
sin
6
5
cos
2
5
cos
6
2
cos ( 2 y − 5 z )
( )
(
))
( )
(
∂y
2

2
−
zx
x
cos
6
2
cos ( 2 y − 5 z )
(
)
∂z
=
2
∂y 6 xy sin ( 6 zx ) − 5 x cos ( 2 y − 5 z )

[Return to Problems]

There’s quite a bit of work to these
...




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