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. Before we actually start
taking derivatives of functions of more than one variable let’s recall an
important interpretation of derivatives of functions of one variable.
Recall that given a function of one variable, ,
the derivative, ,
represents the rate of change of the function as x changes. This is an
important interpretation of derivatives and we are not going to want to lose it
with functions of more than one variable.
The problem with functions of more than one variable is that there is
more than one variable. In other words,
what do we do if we only want one of the variables to change, or if we want
more than one of them to change? In
fact, if we’re going to allow more than one of the variables to change there
are then going to be an infinite amount of ways for them to change. For instance, one variable could be changing
faster than the other variable(s) in the function. Notice as well that it will be completely
possible for the function to be changing differently depending on how we allow
one or more of the variables to change.
We will need to develop ways, and notations, for dealing
with all of these cases. In this section
we are going to concentrate exclusively on only changing one of the variables
at a time, while the remaining variable(s) are held fixed. We will deal with allowing multiple variables
to change in a later section.
Because we are going to only allow one of the variables to
change taking the derivative will now become a fairly simple process. Let’s start off this discussion with a fairly
simple function.
Let’s start with the function and let’s determine the rate at which the
function is changing at a point, ,
if we hold y fixed and allow x to vary and if we hold x fixed and allow y to vary.
We’ll start by looking at the case of holding y fixed and allowing x to vary. Since we are interested in the rate of change
of the function at and are holding y fixed this means that we are going to always have (if we didn’t have this then eventually y would have to change in order to get
to the point…). Doing this will give us
a function involving only x’s and we
can define a new function as follows,
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 at . In other words, we want to compute and since this is a function of a single
variable we already know how to do that.
Here is the rate of change of the function at if we hold y
fixed and allow x to vary.
We will call the partial
derivative of with respect to x at and we will denote it in the following way,
Now, let’s do it the other way. We will now hold x fixed and allow y to
vary. We can do this in a similar
way. Since we are holding x fixed it must be fixed at and so we can define a new function of y and then differentiate this as we’ve
always done with functions of one variable.
Here is the work for this,
In this case we call the partial
derivative of with respect to y at and we denote it as follows,
Note that these two partial derivatives are sometimes called
the first order partial derivatives. Just as with functions of one variable we can
have derivatives of all orders. We will
be looking at higher order derivatives in a later section.
Note that the notation for partial derivatives is different
than that for derivatives of functions of a single variable. With functions of a single variable we could
denote the derivative with a single prime.
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. We will shortly be seeing some alternate
notation for partial derivatives as well.
Note as well that we usually don’t use the notation for partial derivatives. The more standard notation is to just
continue to use . So, the partial derivatives from above will
more commonly be written as,
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.
To compute all we need to do is treat all the y’s as constants (or numbers) and then
differentiate the x’s as we’ve always
done. Likewise, to compute we will treat all the x’s as constants and then differentiate the y’s as we are used to doing.
Before we work any examples let’s get the formal definition
of the partial derivative out of the way as well as some alternate notation.
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. Here are the
formal definitions of the two partial derivatives we looked at
above.
Now let’s take a quick look at some of the possible
alternate notations for partial derivatives.
Given the function the following are all equivalent notations,
For the fractional notation for the partial derivative
notice the difference between the partial derivative and the ordinary
derivative from single variable calculus.
Okay, now let’s work some examples. When working these examples always keep in
mind that we need to pay very close attention to which variable we are
differentiating with respect to. 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. If you can remember this
you’ll find that doing partial derivatives are not much more difficult that
doing derivatives of functions of a single variable as we did in Calculus I.
Example 1 Find
all of the first order partial derivatives for the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
Let’s first take the derivative with respect to x and remember that as we do so all
the y’s will be treated as
constants. The partial derivative with
respect to x is,
Notice that the second and the third term differentiate to
zero in this case. It should be clear
why the third term differentiated to zero.
It’s a constant and we know that constants always differentiate to
zero. This is also the reason that the
second term differentiated to zero.
Remember that since we are differentiating with respect to x here we are going to treat all y’s as constants. That means that terms that only involve y’s will be treated as constants and
hence will differentiate to zero.
Now, let’s take the derivative with respect to y.
In this case we treat all x’s
as constants and so the first term involves only x’s and so will differentiate to zero, just as the third term
will. Here is the partial derivative
with respect to y.
[Return to Problems]
(b)
With this function we’ve got three first order derivatives
to compute. Let’s do the partial
derivative with respect to x
first. Since we are differentiating
with respect to x we will treat all
y’s and all z’s as constants. This
means that the second and fourth terms will differentiate to zero since they
only involve y’s and z’s.
This first term contains both x’s and y’s and so when
we differentiate with respect to x
the y will be thought of as a
multiplicative constant and so the first term will be differentiated just as
the third term will be differentiated.
Here is the partial derivative with respect to x.
Let’s now differentiate with respect to y.
In this case all x’s and z’s will be treated as constants. This means the third term will
differentiate to zero since it contains only x’s while the x’s in
the first term and the z’s in the second
term will be treated as multiplicative constants. Here is the derivative with respect to y.
Finally, let’s get the derivative with respect to z.
Since only one of the terms involve z’s this will be the only nonzero term in the derivative. Also, the y’s in that term will be treated as multiplicative
constants. Here is the derivative with
respect to z.
[Return to Problems]
(c)
With this one we’ll not put in the detail of the first
two. Before taking the derivative
let’s rewrite the function a little to help us with the differentiation
process.
Now, the fact that we’re using s and t here instead of
the “standard” x and y shouldn’t be a problem. It will work the same way. Here are the two derivatives for this
function.
Remember how to differentiate natural logarithms.
[Return to Problems]
(d)
Now, we can’t forget the product rule with
derivatives. The product rule will
work the same way here as it does with functions of one variable. We will just need to be careful to remember
which variable we are differentiating with respect to.
Let’s start out by differentiating with respect to x.
In this case both the cosine and the exponential contain x’s and so we’ve really got a product
of two functions involving x’s and
so we’ll need to product rule this up.
Here is the derivative with respect to x.
Do not forget the chain
rule for functions of one variable.
We will be looking at the chain rule for some more complicated
expressions for multivariable functions in a later section. However, at this point we’re treating all
the y’s as constants and so the
chain rule will continue to work as it did back in Calculus I.
Also, don’t forget how to differentiate exponential
functions,
Now, let’s differentiate with respect to y.
In this case we don’t have a product rule to worry about since the
only place that the y shows up is
in the exponential. 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. Here is the
derivative with respect to y.
[Return to Problems]

Example 2 Find
all of the first order partial derivatives for the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
We also can’t forget about the quotient rule. Since there isn’t too much to this one, we
will simply give the derivatives.
In the case of the derivative with respect to v recall that u’s are constant and so when we differentiate the numerator we
will get zero!
[Return to Problems]
(b)
Now, we do need to be careful however to not use the
quotient rule when it doesn’t need to be used. 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.
Let’s do the derivatives with respect to x and y first. 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. Here are the
derivatives for these two cases.
Now, in the case of differentiation with respect to z we can avoid the quotient rule with
a quick rewrite of the function. Here
is the rewrite as well as the derivative with respect to z.
We went ahead and put the derivative back into the
“original” form just so we could say that we did. In practice you probably don’t really need
to do that.
[Return to Problems]
(c)
In this last part we are just going to do a somewhat messy
chain rule problem. However, if you
had a good background in Calculus I chain
rule this shouldn’t be all that difficult of a problem. Here are the two derivatives,
[Return to Problems]

So, there are some examples of partial derivatives. Hopefully you will agree that as long as we
can remember to treat the other variables as constants these work in exactly
the same manner that derivatives of functions of one variable do. So, if you can do Calculus I derivatives you
shouldn’t have too much difficulty in doing basic partial derivatives.
There is one final topic that we need to take a quick look
at in this section, implicit differentiation.
Before getting into implicit differentiation for multiple variable
functions let’s first remember how implicit differentiation works for functions
of one variable.
Now, we did this problem because implicit differentiation
works in exactly the same manner with functions of multiple variables. 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. In other words, . Then whenever we differentiate z’s with respect to x we will use the chain rule and add on a . Likewise, whenever we differentiate z’s with respect to y we will add on a .
Let’s take a quick look at a couple of implicit
differentiation problems.
There’s quite a bit of work to these. We will see an easier way to do implicit
differentiation in a later section.