Lectures on Calculus

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Lectures on Calculus

• Multivariable Differentiation

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by William M. Faucette

• University of West Georgia

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Adapted from Calculus on Manifolds

• by Michael Spivak

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Multivariable Differentiation

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

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Multivariable Differentiation

• This definition makes no sense for functions
  fRn?Rm for several reasons, not the least of
  which is that you cannot divide by a vector.

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Multivariable Differentiation

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

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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.

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Multivariable Differentiation

• So, we define a function fRn?Rm to be
  differentiable at a in Rn if there exists a
  linear transformation ? from Rn to Rm so that

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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.

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Multivariable Differentiation

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

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Multivariable Differentiation

• Notice that for fRn?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

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Multivariable Differentiation

• For a function fRn?Rm, the derivative Df(a) is
  unique if it exists.
• This result will follow from what we do later.

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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.

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Our First Lemma
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Lemma 1

• Lemma If fRn?Rm is a linear transformation,
  then Df(a)f.

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Lemma 1

• Proof Let ?f. Then

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Our Second Lemma
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Lemma 2

• Lemma Let TRm?Rn be a linear transformation.
  Then there is a number M such that T(h)Mh
  for h2Rm.

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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 Kmaxaijgt0.

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Lemma 2

• Proof Then
• So, we need only let MKm. QED

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The Chain Rule
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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

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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.

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The Chain Rule

• Proof Let bf(a), let ?Df(a), and let
  ?Dg(f(a)). Define

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The Chain Rule

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

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The Chain Rule

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

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The Chain Rule

• To show that g?f is differentiable with
  derivative ???, we must show that

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The Chain Rule

• Recall that
• and that ? is a linear transformation. Then we
  have

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The Chain Rule

• Next, recall that
• Then we have

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The Chain Rule

• From the preceding slide, we have
• So, we must show that

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The Chain Rule

• Recall that
• Given ?gt0, we can find ?gt0 so that
• which is true provided that x-alt?1, since f
  must be continuous at a.

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The Chain Rule

• Then
• Here, weve used Lemma 2 to find M so that

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The Chain Rule

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

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The Chain Rule

• Since ?gt0 is arbitrary, we have
• which is what we needed to show first.

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The Chain Rule

• Recall that
• Given ?gt0, we can find ?2gt0 so that

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The Chain Rule

• By Lemma 2, we can find M so that
• Hence

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The Chain Rule

• Since ?gt0 is arbitrary, we have
• which is what we needed to show second. QED

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The Derivative of fRn?Rm
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The Derivative of fRn?Rm

• Let f be given by m coordinate functions f 1, .
  . . , f m.
• We can first make a reduction to the case where
  m1 using the following theorem.

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The Derivative of fRn?Rm

• Theorem If fRn?Rm, then f is differentiable at
  a2Rn if and only if each f i is differentiable at
  a2Rn, and

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The Derivative of fRn?Rm

• Proof One direction is easy. Suppose f is
  differentiable. Let ?iRm?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).

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The Derivative of fRn?Rm

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

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The Derivative of fRn?Rm

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

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The Derivative of fRn?Rm

• Proof Then
• This concludes the proof. QED

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The Derivative of fRn?Rm

• The preceding theorem reduces differentiating
  fRn?Rm to finding the derivative of each
  component function f iRn?R. Now well work on
  this problem.

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Partial Derivatives
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Partial Derivatives

• Let f Rn?R and a2Rn. We define the ith partial
  derivative of f at a by

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The Derivative of fRn?Rm

• Theorem If fRn?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)).

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The Derivative of fRn?Rm

• Proof Suppose first that m1, so that fRn?R.
  Define hR?Rn by
• h(x)(a1, . . . , x, . . . ,an),
• with x in the jth place. Then

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The Derivative of fRn?Rm

• Proof Hence, by the Chain Rule, we have

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The Derivative of fRn?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

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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.

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When is f differentiable?

• Theorem If fRn?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.)

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When is f differentiable?

• Proof As before, it suffices to consider the
  case when m1, so that fRn?R. Then

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When is f differentiable?

• Proof Applying the Mean Value Theorem, we have
• for some b1 between a1 and a1h1.

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When is f differentiable?

• Proof Applying the Mean Value Theorem in the
  ith place, we have
• for some bi between ai and aihi.

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When is f differentiable?

• Proof Then
• since Dif is continuous at a. QED

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Summary

• We have learned that
• A function fRn ?Rm is differentiable if and only
  if each component function f iRn ?R is
  differentiable

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Summary

• We have learned that
• If fRn ?Rm is differentiable, all the partial
  derivatives of all the component functions exist
  and the matrix Df(a) is given by

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Summary

• We have learned that
• If fRn ?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.