Lectures on Calculus

1. Lectures on Calculus Multivariable Differentiation

2. 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 fi 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 fi= 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 fi= 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 fi is differentiable at a with derivative Dfi(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 fi: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