Lectures on Calculus

Lectures on Calculus

Multivariable Differentiation

by William M. Faucette

University of West Georgia

Adapted from Calculus on Manifolds

by Michael Spivak

Multivariable Differentiation

Recall that a function f: RR is differentiable at a in R if there is a number f (a) such that

Multivariable Differentiation

This definition makes no sense for functions f:RnRm for several reasons, not the least of which is that you cannot divide by a vector.

Multivariable Differentiation

However, we can rewrite this definition so that it can be generalized to several variables. First, rewrite the definition this way

Multivariable Differentiation

Notice that the function taking h to f (a)h is a linear transformation from R to R. So we can view f (a) as being a linear transformation, at least in the one dimensional case.

Multivariable Differentiation

So, we define a function f:RnRm to be differentiable at a in Rn if there exists a linear transformation from Rn to Rm so that

Multivariable Differentiation

Notice that taking the length here is essential since the numerator is a vector in Rm and denominator is a vector in Rn.

Multivariable Differentiation

Definition: The linear transformation is denoted Df(a) and called the derivative of f at a, provided

Multivariable Differentiation

Notice that for f:RnRm, the derivative Df(a):RnRm is a linear transformation. Df(a) is the linear transformation most closely approximating the map f at a, in the sense that

Multivariable Differentiation

For a function f:RnRm, the derivative Df(a) is unique if it exists.

This result will follow from what we do later.

Multivariable Differentiation

Since Df(a) is a linear transformation, we can give its matrix with respect to the standard bases on Rn and Rm. This matrix is an mxn matrix called the Jacobian matrix of f at a.

We will see how to compute this matrix shortly.

Our First Lemma

Lemma 1

Lemma: If f:RnRm is a linear transformation, then Df(a)=f.

Lemma 1

Proof: Let =f. Then

Our Second Lemma

Lemma 2

Lemma: Let T:RmRn be a linear transformation. Then there is a number M such that |T(h)|≤M|h| for h2Rm.

Lemma 2

Proof: Let A be the matrix of T with respect to the standard bases for Rm and Rn. So A is an nxm matrix [aij]

If A is the zero matrix, then T is the zero linear transformation and there is nothing to prove. So assume A≠0.

Let K=max{|aij|}>0.

Lemma 2

Proof: Then

So, we need only let M=Km. QED

The Chain Rule

The Chain Rule

Theorem (Chain Rule): If f: RnRm is differentiable at a, and g: RmRp is differentiable at f(a), then the composition gf: RnRp is differentiable at a and

The Chain Rule

In this expression, the right side is the composition of linear transformations, which, of course, corresponds to the product of the corresponding Jacobians at the respective points.

The Chain Rule

Proof: Let b=f(a), let =Df(a), and let =Dg(f(a)). Define

The Chain Rule

Since f is differentiable at a, and is the derivative of f at a, we have

The Chain Rule

Similarly, since g is differentiable at b, and is the derivative of g at b, we have

The Chain Rule

To show that gf is differentiable with derivative , we must show that

The Chain Rule

Recall that

and that is a linear transformation. Then we have

The Chain Rule

Next, recall that

Then we have

The Chain Rule

From the preceding slide, we have

So, we must show that

The Chain Rule

Recall that

Given >0, we can find >0 so that

which is true provided that |x-a|<1, since f must be continuous at a.

The Chain Rule

Then

Here, we’ve used Lemma 2 to find M so that

The Chain Rule

Dividing by |x-a| and taking a limit, we get

The Chain Rule

Since >0 is arbitrary, we have

which is what we needed to show first.

The Chain Rule

Recall that

Given >0, we can find 2>0 so that

The Chain Rule

By Lemma 2, we can find M so that

Hence

The Chain Rule

Since >0 is arbitrary, we have

which is what we needed to show second. QED

The Derivative of f:RnRm

The Derivative of f:RnRm

Let f be given by m coordinate functions f 1, . . . , f m.

We can first make a reduction to the case where m=1 using the following theorem.

The Derivative of f:RnRm

Theorem: If f:RnRm, then f is differentiable at a2Rn if and only if each f i is differentiable at a2Rn, and

The Derivative of f:RnRm

Proof: One direction is easy. Suppose f is differentiable. Let i:RmR be projection onto the ith coordinate. Then f i= if. Since i is a linear transformation, by Lemma 1 it is differentiable and is its own derivative. Hence, by the Chain Rule, we have f i= if is differentiable and Df i(a) is the ith component of Df(a).

The Derivative of f:RnRm

Proof: Conversely, suppose each f i is differentiable at a with derivative Df i(a).

Set

Then

The Derivative of f:RnRm

Proof: By the definition of the derivative, we have, for each i,

The Derivative of f:RnRm

Proof: Then

This concludes the proof. QED

The Derivative of f:RnRm

The preceding theorem reduces differentiating f:RnRm to finding the derivative of each component function f i:RnR. Now we’ll work on this problem.

Partial Derivatives

Partial Derivatives

Let f: RnR and a2Rn. We define the ith partial derivative of f at a by

The Derivative of f:RnRm

Theorem: If f:RnRm is differentiable at a, then Djf i(a) exists for 1≤ i ≤m, 1≤ j ≤n and f(a) is the mxn matrix (Djf i(a)).

The Derivative of f:RnRm

Proof: Suppose first that m=1, so that f:RnR. Define h:RRn by

h(x)=(a1, . . . , x, . . . ,an),

with x in the jth place. Then

The Derivative of f:RnRm

Proof: Hence, by the Chain Rule, we have

The Derivative of f:RnRm

Proof: Since (fh)(aj) has the single entry Djf(a), this shows that Djf(a) exists and is the jth entry of the 1xn matrix f (a).

The theorem now follows for arbitrary m since, by our previous theorem, each f i is differentiable and the ith row of f (a) is (f i)(a). QED

Pause

Now we know that a function f is differentiable if and only if each component function f i is and that if f is differentiable, Df(a) is given by the matrix of partial derivatives of the component functions f i.

What we need is a condition to ensure that f is differentiable.

When is f differentiable?

Theorem: If f:RnRm, then Df(a) exists if all Djf i(x) exist in an open set containing a and if each function Djf i is continuous at a.

(Such a function f is called continuously differentiable.)

When is f differentiable?

Proof: As before, it suffices to consider the case when m=1, so that f:RnR. Then

When is f differentiable?

Proof: Applying the Mean Value Theorem, we have

for some b1 between a1 and a1+h1.

When is f differentiable?

Proof: Applying the Mean Value Theorem in the ith place, we have

for some bi between ai and ai+hi.

When is f differentiable?

Proof: Then

since Dif is continuous at a. QED

Summary

We have learned that • A function f:Rn Rm is differentiable if and

only if each component function f i:Rn R is differentiable;

Summary

We have learned that • If f:Rn Rm is differentiable, all the partial

derivatives of all the component functions exist and the matrix Df(a) is given by

Summary

We have learned that • If f:Rn Rm and all the partial derivatives

Djf i(a) exist in a neighborhood of a and are continuous at a, then f is differentiable at a.