Given a function , the change from to is . For example, if , then the change in going from to is
This can be seen as a linear function plus another part which is small in the sense that
Definition 1: If is a function, then a derivative of at is a linear function from to such that
Proposition 1: There is at most one derivative of at .
Proof: Suppose that and are two derivatives. Then each satisfies the limit condition of the definition. Subtracting these gives:
For , let for . Then
since and are linear. So
Exercise 1: If is linear, then for all . If is a constant function, then for all .
Proposition 2: If and , then is differentiable at if and only if the coordinate functions are differentiable for all . If either is true, then .
Proof: If is differentiable at , then
In particular each of the terms of the sum have limit zero. So the coordinate functions have derivatives and they are the coordinate functions of the derivative of . Conversely, if the coordinate functions are differentiable, then putting their derivatives in the above formula in the place of the coordinate functions of shows that is differentiable at .
The main computational tool for derivatives is:
Theorem 1: (Chain Rule) If is differentiable at and if is differentiable at , then is differentiable at and .
Proof: Define and by
We need to show that where
Now, letting and , we have by the definition of ,
Using the definition of , we get
By linearity of , this simplifies to:
We must show that . Since is linear, there is an such that
and the right hand side approaches zero.
For the other term, for any , there is a such that
whenever . Note that we can remove the condition that the middle term be non-zero -- if it were zero, then the condition becomes which is true since . Now is bounded for small . So
This completes the proof.
Corollary 1: If are differentiable at , then
Proof: The sum (respectively the product ) function can be written as the composition of the defined by and the function (respectively ) defined by (respectively ). Now is linear and so its derivative is itself. Further, it is easy to check that:
and so since
The formulas now follow by the chain rule.
Definition 2: Let be a function and . Then, for each j, one can define by . The partial derivative of at is defined to be .
Proposition 3: Let be a function with domain and be in the interior of and assume that is either a local minimum or local maximum of . Then provided that the partial derivative exists.
Proof: is a local extremum of the function of one variable .
Theorem 2: Let be a function and . If is differentiable at , then is the linear transformation whose matrix is the array of partial derivatives of the coordinate functions of . Conversely, if The partial derivatives of the coordinate functions of all have partials defined in an open neighborhood of and are continuous at , then is differentiable at .
Proof: By Proposition 2, it is enough to check the case where . The first assertion is immediate from the definitions. For the second assertion break up the change in into a sum of changes in which one varies only one parameter at a time:
One can apply the Mean Value Theorem to each of the differences on the right:
where . It follows that
because the are continuous at .
Corollary 2: (Chain Rule) If for are continuously differentiable at and if is differentiable at (g_1(a),...,g_m(a)), then has partial derivatives
Proof: Apply the Chain Rule and Theorem 2.
Theorem 3: If is a function defined on an open subset and if and its first and second order partials are exist throughout and are continuous there, then for all .
Proof: It is clearly enough to prove the result in the case where . Let and be such that the rectangle with diagonal from to is contained in . Let be defined by . First apply the Mean Value Theorem to find a such that
Now, . Apply the Mean Value Theorem again to find between and such that
Combining results we have
which is symmetric in the two variables. So, one can obtain another expression for the same by repeating the same construction swapping the roles of the two variables: Let and get and such that
So . Now take the limit as and use the continuity of the second partials to conclude that the mixed partials are equal.