## 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**

### 11.2: 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 point**a ∈ R**^{n}if and only
if there is an open set V containing**a such that f : V → R**^{m}
and there is a T ∈ L(R^{n};**R**^{m})such that the function

ε(h) := f (a + h) − f (a) − T (h)

(defined for**h 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)**

Theorem (11.13)

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

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

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

Proof.

Suppose that f is differentiable at**a.**Then by
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

W -C L **Advanced Calculus (II)**

Theorem (11.13)

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

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

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

W -C L **Advanced Calculus (II)**

Theorem (11.13)

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

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

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

W -C L **Advanced Calculus (II)**

Theorem (11.13)

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

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

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

Proof.

Suppose that f is differentiable at**a. Then by**
lim**h→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)
as**h → 0; i.e., f is continuous at a.**

W -C L **Advanced Calculus (II)**

Theorem (11.14)

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

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

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

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

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

W -C L **Advanced Calculus (II)**

Theorem (11.14)

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

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

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

**a, then all**
first-order partial derivatives of f exists at**a.**

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

W -C L **Advanced Calculus (II)**

Theorem (11.14)

**a, then all**
first-order partial derivatives of f exists at**a.**

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

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

**a, then all**
first-order partial derivatives of f exists at**a.**

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

W -C L **Advanced Calculus (II)**

Theorem (11.14)

**a, then all**
first-order partial derivatives of f exists at**a.**

Proof.

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

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

f (a + h) − f (a) − Bh

**khk** := f (a + tej) −f (a)

t **− Be**_{j}.

Take the limit of this identity as t → 0+, using
lim**h→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 =Be_{j} = (b_{1j}, . . . ,b_{mj}).

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

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 at**a and satisfies**

∂fi

∂xj

(a) = bij

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

(4) B = ∂fi

∂xj

(a)

m×n

:=

∂f1

∂x1(a) · · · _{∂x}^{∂f}^{1}

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

∂fm

∂x_{1}(a) · · · ^{∂f}_{∂x}^{m}

n(a)

.

W -C L **Advanced Calculus (II)**

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 at**a and satisfies**

∂fi

∂xj

(a) = bij

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

(4) B = ∂fi

∂xj

(a)

m×n

:=

∂f1

∂x1(a) · · · _{∂x}^{∂f}^{1}

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

∂fm

∂x_{1}(a) · · · ^{∂f}_{∂x}^{m}

n(a)

.

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

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 at**a and satisfies**

∂fi

∂xj

(a) = bij

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

(4) B = ∂fi

∂xj

(a)

m×n

:=

∂f1

∂x1(a) · · · _{∂x}^{∂f}^{1}

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

∂fm

∂x_{1}(a) · · · ^{∂f}_{∂x}^{m}

n(a)

.

W -C L **Advanced Calculus (II)**

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 at**a and satisfies**

∂fi

∂xj

(a) = bij

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

(4) B = ∂fi

∂xj

(a)

m×n

:=

∂f1

∂x1(a) · · · _{∂x}^{∂f}^{1}

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

∂fm

∂x_{1}(a) · · · ^{∂f}_{∂x}^{m}

n(a)

.

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

Df (a) := [∂fi

∂xj

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

W -C L **Advanced Calculus (II)**

Theorem (11.15)

Let V be open in**R**^{n}, let**a ∈ V , and suppose that**

f : V →**R**^{m}. If all first-order partial derivatives of f exist in
V and are continuous at**a, then f is differentiable at a.**

Note: These hypotheses are met if f is C^{1}on 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→0**lim

f (a + h) − f (a) − ∇f (a) · h

**khk** =0.

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

Let V be open in**R**^{n}, let**a ∈ V , and suppose that**

f : V →**R**^{m}. If all first-order partial derivatives of f exist in
V and are continuous at**a, then f is differentiable at a.**

Note: These hypotheses are met if f is C^{1}on 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→0**lim

f (a + h) − f (a) − ∇f (a) · h

**khk** =0.

W -C L **Advanced Calculus (II)**

Theorem (11.15)

Let V be open in**R**^{n}, let**a ∈ V , and suppose that**

f : V →**R**^{m}. If all first-order partial derivatives of f exist in
V and are continuous at**a, then f is differentiable at a.**

Note: These hypotheses are met if f is C^{1}on 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→0**lim

f (a + h) − f (a) − ∇f (a) · h

**khk** =0.

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

Let V be open in**R**^{n}, let**a ∈ V , and suppose that**

**R**^{m}. If all first-order partial derivatives of f exist in
V and are continuous at**a, then f is differentiable at a.**

Note: These hypotheses are met if f is C^{1}on 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→0**lim

f (a + h) − f (a) − ∇f (a) · h

**khk** =0.

W -C L **Advanced Calculus (II)**

Proof.

Let**a = (a**1, . . . ,an).Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(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,

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

Let**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V .Fix**h = (h**_{1}, . . . ,hn) 6=**0 in B**_{r}(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)**

Proof.

Let**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(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,

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

Let**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(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)**

Proof.

Let**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(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,

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

Let**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(0). By

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)**

Proof.

**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(0). By

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,

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

**a = (a**1, . . . ,an). Choose r > 0 so small that
Br(a) ⊂ V . Fix h = (h_{1}, . . . ,hn) 6=**0 in B**_{r}(0). By

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)**

Proof.

where δ ∈**R**^{n} 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
at**a, δ**j **→ 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 as**h → 0. Thus f is differentiable at a**
by definition.

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

where δ ∈**R**^{n} 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
at**a, δ**j → 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 as**h → 0. Thus f is differentiable at a**
by definition.

W -C L **Advanced Calculus (II)**

Proof.

where δ ∈**R**^{n} 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
at**a, δ**j **→ 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 as**h → 0. Thus f is differentiable at a**
by definition.

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

where δ ∈**R**^{n} 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
at**a, δ**j **→ 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.

**h → 0.**Thus f is differentiable at**a**
by definition.

W -C L **Advanced Calculus (II)**

Proof.

where δ ∈**R**^{n} is the vector with components
δj = ∂f

∂xj

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

∂xj

(a1, . . . ,an).

**a, δ**j **→ 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.

**h → 0. Thus f is differentiable at a**
by definition.

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

where δ ∈**R**^{n} is the vector with components
δj = ∂f

∂xj

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

∂xj

(a1, . . . ,an).

**a, δ**j **→ 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.

**h → 0.**Thus f is differentiable at**a**
by definition.

W -C L **Advanced Calculus (II)**

Proof.

where δ ∈**R**^{n} is the vector with components
δj = ∂f

∂xj

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

∂xj

(a1, . . . ,an).

**a, δ**j **→ 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.

**h → 0. Thus f is differentiable at a**
by definition.

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

Prove that

f (x , y ) =( (x^{2}+y^{2})sin√ ^{1}

x^{2}+y^{2} (x , y ) 6= (0, 0)

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

is differentiable on**R**^{2}but not continuously differentiable at
(0, 0).

W -C L **Advanced Calculus (II)**

## Thank you.

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