State the chain rules for one or two independent variables.
Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables.
Perform implicit differentiation of a function of two or more variables.
In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. In this section, we study extensions of the chain rule and learn how to take derivatives of compositions of functions of more than one variable.
Chain rules for one or two independent variables
Recall that the chain rule for the derivative of a composite of two functions can be written in the form
In this equation, both
and
are functions of one variable. Now suppose that
is a function of two variables and
is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.
Chain rule for one independent variable
Suppose that
and
are differentiable functions of
and
is a differentiable function of
Then
is a differentiable function of
and
where the ordinary derivatives are evaluated at
and the partial derivatives are evaluated at
Proof
The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that
f is differentiable at the point
where
and
for a fixed value of
We wish to prove that
is differentiable at
and that
[link] holds at that point as well.
Since
is differentiable at
we know that
where
We then subtract
from both sides of this equation:
Next, we divide both sides by
Then we take the limit as
approaches
The left-hand side of this equation is equal to
which leads to
The last term can be rewritten as
As
approaches
approaches
so we can rewrite the last product as
Since the first limit is equal to zero, we need only show that the second limit is finite:
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