lectures on calculus
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Lectures on Calculus. Multivariable Differentiation. by William M. Faucette. University of West Georgia. Adapted from Calculus on Manifolds. by Michael Spivak. Multivariable Differentiation. - PowerPoint PPT PresentationTRANSCRIPT
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.