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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch11: Differentiability on R

### 11.4: Chain Rule

Theorem (11.28 Chain Rule)

Suppose thata ∈ Rn, that g is a vector function from n variables to m variables, and that f is a vector function from m variables to p variables. If g is differentiable ata and f is differentiable at g(a), then f ◦ g is differentiable at a and

(20) D(f ◦ g)(a) = Df (g(a))Dg(a).

(The product Df (g(a))Dg(a) is a matrix multiplication.)

(3)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g.By the

uniqueness of the total derivative, we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(4)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the

uniqueness of the total derivative,we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(5)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g.By the

uniqueness of the total derivative, we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(6)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the

uniqueness of the total derivative,we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(7)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the

uniqueness of the total derivative, we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(8)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the

uniqueness of the total derivative, we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(9)

Set T = Df (g(a))Dg(a) and observe that T , the product of a p × m matrix with an m × n matrix, is a p × n matrix, the right size for the total derivative of f ◦ g. By the

uniqueness of the total derivative, we must show that

(21) lim

h→0

f (g(a + h)) − f (g(a)) − T (h)

khk =0.

Letb = g(a). Set

(22) ε(h) = g(a + h) − g(a) − Dg(a)(h) and

(23) δ(k) = f (b + k) − f (b) − Df (b)(k) forh and k sufficiently small.

(10)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn,and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a).Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(11)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm.Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a))=f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(12)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a).Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b)=Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(13)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a))=f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(14)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b)=Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k)=T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(15)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(16)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k)=T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h)=Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(17)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(18)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h)=Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(19)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(20)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(21)

By hypothesis, ε(h)khk → 0 in Rm ash → 0 in Rn, and

δ(k)

kkk → 0 in Rp ask → 0 in Rm. Fixh small and set k = g(a + h) − g(a). Since (23) and (22) imply

f (g(a + h)) − f (g(a)) = f (b + k) − f (b) = Df (b)(k) + δ(k)

=Df (b)(Dg(a)(h)+ε(h))+δ(k) = T (h)+Df (b)(ε(h))+δ(k), we have

f (g(a + h)) − f (g(a)) − T (h) = Df (b)(ε(h)) + δ(k)

=:T1(h) + T2(h).

It remains to verify that Tkhkj(h) → 0 as h → 0 for j = 1, 2.

(22)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication,it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm.we have

kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(23)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0.On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have

kkk := kg(a + h) − g(a)k= kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(24)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm.we have

kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(25)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have

kkk := kg(a + h) − g(a)k= kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(26)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have

kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(27)

Since ε(h)khk → 0 as h → 0 and Df (b)(h) is matrix multiplication, it is clear that Tkhk1(h) → Df (b)(0) = 0 as h → 0. On the other hand, by (22), the triangle inequality, and the definition of the operator norm. we have

kkk := kg(a + h) − g(a)k = kDg(a)(h) + ε(h)k

≤ kDg(a)k · khk + kε(h)k.

(28)

Thus kkkkhk is bounded forh sufficiently small.Sincek → 0 inRm ash → 0 in Rn, it follows that

kT2(h)k

khk = kkk

khk· kδ(k)k kkk → 0

ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).

(29)

Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn,it follows that

kT2(h)k

khk = kkk

khk· kδ(k)k kkk → 0

ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).

(30)

Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that

kT2(h)k

khk = kkk

khk· kδ(k)k kkk → 0

ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).

(31)

Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that

kT2(h)k

khk = kkk

khk· kδ(k)k kkk → 0

ash → 0.We conclude that f ◦ g is differentiable ata and the differentiable ata and the derivative is Df (g(a))Dg(a).

(32)

Thus kkkkhk is bounded forh sufficiently small. Since k → 0 inRm ash → 0 in Rn, it follows that

kT2(h)k

khk = kkk

khk· kδ(k)k kkk → 0

ash → 0. We conclude that f ◦ g is differentiable at a and the differentiable ata and the derivative is Df (g(a))Dg(a).

(33)

Let f :R2→ R be C2 onR2and set u(r , θ)

=f (r cos θ, r sin θ). If f satisfies the Laplace equation, i.e., if

2f

∂x2 + ∂2f

∂y2 =0 prove for each r 6= 0 that

1 r2

2u

∂θ2 +1 r

∂u

∂r + ∂2u

∂r2 =0.

(34)

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