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(1)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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Ch11: Differentiability on R

11.2: Definition of Differentiability

Definition (11.12)

Let f be a vector function from n variables to m variables.

(i) f is said to be differentiable at a pointa ∈ Rnif and only if there is an open set V containinga such that f : V → Rm and there is a T ∈ L(Rn;Rm)such that the function

ε(h) := f (a + h) − f (a) − T (h)

(defined forh sufficiently small) satisfies ε(h)khk → 0 as h → 0.

(ii) f is said to be differentiable on a set E if and only if E is nonempty, and f is differentiable at every point in E .

W -C L Advanced Calculus (II)

(3)

Theorem (11.13)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0,there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

(4)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata.Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ.By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

W -C L Advanced Calculus (II)

(5)

Theorem (11.13)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0,there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm,it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

(6)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ.By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ.Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

W -C L Advanced Calculus (II)

(7)

Theorem (11.13)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm,it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number,we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

(8)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ.Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

W -C L Advanced Calculus (II)

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Theorem (11.13)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number,we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

(10)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

W -C L Advanced Calculus (II)

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Theorem (11.14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0.Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

(12)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0.Fix 1 ≤ j ≤ n and seth = tej for some t > 0. Since khk = t,we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

W -C L Advanced Calculus (II)

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Theorem (11.14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0.Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

(14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t,we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication.We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

W -C L Advanced Calculus (II)

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Theorem (11.14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

(16)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication.We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

W -C L Advanced Calculus (II)

(17)

Theorem (11.14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Bej.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Bej = (b1j, . . . ,bmj).

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A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj).Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

first-order partial derivative of each component fi with respect to xj exists ata and satisfies

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m.In particular,

(4) B = ∂fi

∂xj

(a)



m×n

:=

∂f1

∂x1(a) · · · ∂x∂f1

n(a) ... . .. ...

∂fm

∂x1(a) · · · ∂f∂xm

n(a)

.

W -C L Advanced Calculus (II)

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Proof.

A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15),it follows that the

first-order partial derivative of each component fi with respect to xj exists ata and satisfies

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m. In particular,

(4) B = ∂fi

∂xj

(a)



m×n

:=

∂f1

∂x1(a) · · · ∂x∂f1

n(a) ... . .. ...

∂fm

∂x1(a) · · · ∂f∂xm

n(a)

.

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A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

first-order partial derivative of each component fi with respect to xj exists ata and satisfies

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m.In particular,

(4) B = ∂fi

∂xj

(a)



m×n

:=

∂f1

∂x1(a) · · · ∂x∂f1

n(a) ... . .. ...

∂fm

∂x1(a) · · · ∂f∂xm

n(a)

.

W -C L Advanced Calculus (II)

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Proof.

A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

first-order partial derivative of each component fi with respect to xj exists ata and satisfies

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m. In particular,

(4) B = ∂fi

∂xj

(a)



m×n

:=

∂f1

∂x1(a) · · · ∂x∂f1

n(a) ... . .. ...

∂fm

∂x1(a) · · · ∂f∂xm

n(a)

.

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Df (a) := [∂fi

∂xj

(a)]m×n, i = 1, 2, . . . , m j = 1, 2, . . . , n

W -C L Advanced Calculus (II)

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Theorem (11.15)

Let V be open inRn, leta ∈ V , and suppose that

f : V →Rm. If all first-order partial derivatives of f exist in V and are continuous ata, then f is differentiable at a.

Note: These hypotheses are met if f is C1on V . Proof.

Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition,then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

(24)

Let V be open inRn, leta ∈ V , and suppose that

f : V →Rm. If all first-order partial derivatives of f exist in V and are continuous ata, then f is differentiable at a.

Note: These hypotheses are met if f is C1on V . Proof.

Since a function converges if and only if its components converge (see Theorem 9.15),we may suppose that m = 1. By definition, then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

W -C L Advanced Calculus (II)

(25)

Theorem (11.15)

Let V be open inRn, leta ∈ V , and suppose that

f : V →Rm. If all first-order partial derivatives of f exist in V and are continuous ata, then f is differentiable at a.

Note: These hypotheses are met if f is C1on V . Proof.

Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition,then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

(26)

Let V be open inRn, leta ∈ V , and suppose that

f : V →Rm. If all first-order partial derivatives of f exist in V and are continuous ata, then f is differentiable at a.

Note: These hypotheses are met if f is C1on V . Proof.

Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition, then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

W -C L Advanced Calculus (II)

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Proof.

Leta = (a1, . . . ,an).Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0).By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

(28)

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V .Fixh = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem,we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

W -C L Advanced Calculus (II)

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Proof.

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0).By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a)=f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

(30)

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem,we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

W -C L Advanced Calculus (II)

(31)

Proof.

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a)=f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

(32)

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

W -C L Advanced Calculus (II)

(33)

Proof.

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

(34)

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h1, . . . ,hn) 6=0 in Br(0). By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

W -C L Advanced Calculus (II)

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Proof.

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5),

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

(36)

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n;i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5),

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

W -C L Advanced Calculus (II)

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Proof.

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5),

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

(38)

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5), (6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0.Thus f is differentiable ata by definition.

W -C L Advanced Calculus (II)

(39)

Proof.

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5), (6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

(40)

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5), (6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0.Thus f is differentiable ata by definition.

W -C L Advanced Calculus (II)

(41)

Proof.

where δ ∈Rn is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5), (6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

(42)

Prove that

f (x , y ) =( (x2+y2)sin√ 1

x2+y2 (x , y ) 6= (0, 0)

0 (x , y ) = (0, 0)

is differentiable onR2but not continuously differentiable at (0, 0).

W -C L Advanced Calculus (II)

(43)

Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung