1. Multivariable Differentiation
Lectures on Calculus
Multivariable Differentiation

2. University of West Georgia
by William M. Faucette
University of West Georgia

3. Adapted from Calculus on Manifolds
by Michael Spivak

4. Multivariable Differentiation
Recall that a function f: RR is differentiable at a in R if there is a number f (a) such that

5. Multivariable Differentiation
This definition makes no sense for functions f:RnRm for several reasons, not the least of which is that you cannot divide by a vector.

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

7. 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.

8. Multivariable Differentiation
So, we define a function f:RnRm to be differentiable at a in Rn if there exists a linear transformation  from Rn to Rm so that

9. 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.

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

11. Multivariable Differentiation
Notice that for f:RnRm, the derivative Df(a):RnRm is a linear transformation. Df(a) is the linear transformation most closely approximating the map f at a, in the sense that

12. Multivariable Differentiation
For a function f:RnRm, the derivative Df(a) is unique if it exists.
This result will follow from what we do later.

13. 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.

14. Our First Lemma

15. Lemma 1
Lemma: If f:RnRm is a linear transformation, then Df(a)=f.

16. Lemma 1
Proof: Let =f. Then

17. Our Second Lemma

18. Lemma 2
Lemma: Let T:RmRn be a linear transformation. Then there is a number M such that |T(h)|≤M|h| for h2Rm.

19. 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.

20. Lemma 2
Proof: Then
So, we need only let M=Km. QED

21. The Chain Rule

22. The Chain Rule
Theorem (Chain Rule): If f: RnRm is differentiable at a, and g: RmRp is differentiable at f(a), then the composition gf: RnRp is differentiable at a and

23. 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.

24. The Chain Rule
Proof: Let b=f(a), let =Df(a), and let =Dg(f(a)). Define

25. The Chain Rule
Since f is differentiable at a, and  is the derivative of f at a, we have

26. The Chain Rule
Similarly, since g is differentiable at b, and  is the derivative of g at b, we have

27. The Chain Rule
To show that gf is differentiable with derivative , we must show that

28. The Chain Rule Recall that
and that  is a linear transformation. Then we have

29. The Chain Rule
Next, recall that
Then we have

30. The Chain Rule
From the preceding slide, we have
So, we must show that

31. 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.

32. The Chain Rule
Then
Here, we’ve used Lemma 2 to find M so that

33. The Chain Rule
Dividing by |x-a| and taking a limit, we get

34. The Chain Rule Since >0 is arbitrary, we have
which is what we needed to show first.

35. The Chain Rule
Recall that
Given >0, we can find 2>0 so that

36. The Chain Rule
By Lemma 2, we can find M so that
Hence

37. The Chain Rule Since >0 is arbitrary, we have
which is what we needed to show second. QED

38. The Derivative of f:RnRm

39. The Derivative of f:RnRm
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.

40. The Derivative of f:RnRm
Theorem: If f:RnRm, then f is differentiable at a2Rn if and only if each f i is differentiable at a2Rn, and

41. The Derivative of f:RnRm
Proof: One direction is easy. Suppose f is differentiable. Let i:RmR be projection onto the ith coordinate. Then f i= if. 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= if is differentiable and Df i(a) is the ith component of Df(a).

42. The Derivative of f:RnRm
Proof: Conversely, suppose each f i is differentiable at a with derivative Df i(a).
Set
Then

43. The Derivative of f:RnRm
Proof: By the definition of the derivative, we have, for each i,

44. The Derivative of f:RnRm
Proof: Then
This concludes the proof. QED

45. The Derivative of f:RnRm
The preceding theorem reduces differentiating f:RnRm to finding the derivative of each component function f i:RnR. Now we’ll work on this problem.

46. Partial Derivatives

47. Partial Derivatives
Let f: RnR and a2Rn. We define the ith partial derivative of f at a by

48. The Derivative of f:RnRm
Theorem: If f:RnRm 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)).

49. The Derivative of f:RnRm
Proof: Suppose first that m=1, so that f:RnR. Define h:RRn by
h(x)=(a1, , x, ,an),
with x in the jth place. Then

50. The Derivative of f:RnRm
Proof: Hence, by the Chain Rule, we have

51. The Derivative of f:RnRm
Proof: Since (fh)(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

52. 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.

53. When is f differentiable?
Theorem: If f:RnRm, 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.)

54. When is f differentiable?
Proof: As before, it suffices to consider the case when m=1, so that f:RnR. Then

55. When is f differentiable?
Proof: Applying the Mean Value Theorem, we have
for some b1 between a1 and a1+h1.

56. When is f differentiable?
Proof: Applying the Mean Value Theorem in the ith place, we have
for some bi between ai and ai+hi.

57. When is f differentiable?
Proof: Then
since Dif is continuous at a. QED

58. 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;

59. 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

60. 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.