## Advanced Calculus (II)

W^{EN}-C^{HING}L^{IEN}

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN **Advanced Calculus (II)**

## Ch11: Differentiability on **R**

### 11.4: Chain Rule

Theorem (11.28 Chain Rule)

Suppose that**a ∈ R**^{n}, 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 at**a**
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.)

WEN-CHINGLIEN **Advanced Calculus (II)**

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.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

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.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

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.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

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.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

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.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

uniqueness of the total derivative, we must show that

(21) lim

**h→0**

f (g(a + h)) − f (g(a)) − T (h)

**khk** =0.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

uniqueness of the total derivative, we must show that

(21) lim

**h→0**

f (g(a + h)) − f (g(a)) − T (h)

**khk** =0.

Let**b = 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)
for**h and k sufficiently small.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n},and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}.Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

By hypothesis, ^{ε(h)}_{khk}**→ 0 in R**^{m} as**h → 0 in R**^{n}, and

δ(k)

**kkk** **→ 0 in R**^{p} as**k → 0 in R**^{m}. Fix**h 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 ^{T}_{khk}^{j}^{(h)} **→ 0 as h → 0 for j = 1, 2.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Since ^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication,it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Since ^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication, it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Since ^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication, it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Since ^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication, it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Since ^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication, it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

^{ε(h)}_{khk}**→ 0 as h → 0 and Df (b)(h) is matrix**
multiplication, it is clear that ^{T}_{khk}^{1}^{(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.**

WEN-CHINGLIEN **Advanced Calculus (II)**

Thus ^{kkk}** _{khk}** is bounded for

**h sufficiently small.**Since

**k → 0**in

**R**

^{m}as

**h → 0 in R**

^{n}, it follows that

kT2(h)k

**khk** = **kkk**

**khk**· **kδ(k)k**
**kkk** **→ 0**

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

WEN-CHINGLIEN **Advanced Calculus (II)**

Thus ^{kkk}** _{khk}** is bounded for

**h sufficiently small. Since k → 0**in

**R**

^{m}as

**h → 0 in R**

^{n},it follows that

kT2(h)k

**khk** = **kkk**

**khk**· **kδ(k)k**
**kkk** **→ 0**

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

WEN-CHINGLIEN **Advanced Calculus (II)**

Thus ^{kkk}** _{khk}** is bounded for

**h sufficiently small. Since k → 0**in

**R**

^{m}as

**h → 0 in R**

^{n}, it follows that

kT2(h)k

**khk** = **kkk**

**khk**· **kδ(k)k**
**kkk** **→ 0**

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

WEN-CHINGLIEN **Advanced Calculus (II)**

Thus ^{kkk}** _{khk}** is bounded for

**h sufficiently small. Since k → 0**in

**R**

^{m}as

**h → 0 in R**

^{n}, it follows that

kT2(h)k

**khk** = **kkk**

**khk**· **kδ(k)k**
**kkk** **→ 0**

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

WEN-CHINGLIEN **Advanced Calculus (II)**

^{kkk}** _{khk}** is bounded for

**h sufficiently small. Since k → 0**in

**R**

^{m}as

**h → 0 in R**

^{n}, it follows that

kT2(h)k

**khk** = **kkk**

**khk**· **kδ(k)k**
**kkk** **→ 0**

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

WEN-CHINGLIEN **Advanced Calculus (II)**

Let f :**R**^{2}**→ R be C**^{2} on**R**^{2}and set u(r , θ)

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

∂^{2}f

∂x^{2} + ∂^{2}f

∂y^{2} =0
prove for each r 6= 0 that

1
r^{2}

∂^{2}u

∂θ^{2} +1
r

∂u

∂r + ∂^{2}u

∂r^{2} =0.

WEN-CHINGLIEN **Advanced Calculus (II)**

## Thank you.

WEN-CHINGLIEN **Advanced Calculus (II)**