Syllabus (UNIT-I, C9T, SEM - IV)

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables

(UNIT-I, C9T, SEM - IV)

Functions of several variables, limit and continuity of functions of two or more variables, Partial differentiation, total differentiability and differentiability, sufficient condition for differentiability, Chain rule for one and two independent parameters.

Continuity of function of two variables

Definition: Let (a,b) be a point of the domain D⊂R² of a function f(x,y) such that f(a,b) exists finitely. Then the function f(x,y) is said to be continuous at (a,b) if for any

arbitrary 𝜖 > 0, there exists a 𝛿 > 0 such that

|𝑓(𝑥, 𝑦) − 𝑓(𝑎, 𝑏)| < 𝜖, ∀(𝑥, 𝑦) 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 |𝑥 − 𝑎| < 𝛿, |𝑦 − 𝑏| < 𝛿 𝑜𝑟, (𝑥 − 𝑎)²+ (𝑦 − 𝑏)² < 𝛿². i.e., f(x,y) is said to be continuous at the point (a,b) if lim

(𝑥,𝑦)→(𝑎,𝑏)𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏).

Problem: Let 𝒇(𝒙, 𝒚) = {(𝒙^𝟑+ 𝒚^𝟑)(𝒙 − 𝒚) 𝒘𝒉𝒆𝒏 𝒙 ≠ 𝒚 𝒂𝒏𝒅 𝟎 𝒘𝒉𝒆𝒏 𝒙 = 𝒚. Show that f is continuous at origin.

Solution: Let 𝜖 > 0 be given we have to find a 𝛿 > 0 such that in that δ-nbd of (0,0)

∀(𝑥, 𝑦) we have |𝑓(𝑥, 𝑦) − 𝑓(0,0)| < 𝜖.

Now |𝑓(𝑥, 𝑦) − 𝑓(0,0)| = |(𝑥³+ 𝑦³)(𝑥 − 𝑦) − 0| = |((𝑥²− 𝑦²)((𝑥²− 𝑥𝑦 + 𝑦²)|.

Setting 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃 we get

|𝑓(𝑥, 𝑦) − 𝑓(0,0)| = |𝑟⁴(cos²𝜃 − sin²𝜃)(cos²𝜃 − sin 𝜃 cos 𝜃 + sin²𝜃)|

= |𝑟⁴cos 2𝜃(1 −sin 2𝜃

2 )| ≤ |𝑟⁴(1 −sin 2𝜃

2 )| [𝑠𝑖𝑛𝑐𝑒 | cos 2𝜃| ≤ 1]

≤ 𝑟⁴−^𝑟4

2 | sin 2𝜃| [𝑠𝑖𝑛𝑐𝑒 | sin 2𝜃| ≤ 1] ≤ ^𝑟₂⁴ < 𝜖, whenever, ^𝑟₂⁴< 𝜖 ⇒ (𝑥²+ 𝑦²)² < 2𝜖 ⇒ (𝑥²+ 𝑦²) < 𝛿², 𝛿 = √2𝜖.

Therefore, ∀𝜖 > 0, ∃ 𝛿 > 0 such that

|𝑓(𝑥, 𝑦) − 𝑓(0,0)| < 𝜖, ∀(𝑥, 𝑦) 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 (𝑥²+ 𝑦²) < 𝛿², 𝛿 = √2𝜖.

Therefore, lim

(𝑥,𝑦)→(0,0)𝑓(𝑥, 𝑦) = 𝑓(0,0). So f is continuous at origin.

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables Higher Order Partial Derivatives:

Let f(x,y) be a two variable function in x and y defined in some domain D⊂R² . Then we define the partial derivative with respect to x at (a,b) by [^𝜹𝒇

𝜹𝒙]

(𝒂,𝒃) 𝒐𝒓 𝒇_𝒙(𝒂, 𝒃), defined by

[

^𝜹𝒇_𝜹𝒙

]

(𝒂,𝒃)

= 𝐥𝐢𝐦

𝒉→𝟎

𝒇(𝒂+𝒉,𝒃)−𝒇(𝒂,𝒃)

𝒉

(if this limit exists).

Similarly, [

^𝜹𝒇_𝜹𝒚

]

(𝒂,𝒃)

= 𝐥𝐢𝐦

𝒌→𝟎

𝒇(𝒂,𝒃+𝒌)−𝒇(𝒂,𝒃)

𝒌

(if this limit exists).

Problem: Let 𝒇(𝒙, 𝒚) = ^{𝒙𝟒+𝒚𝟒}

𝒙−𝒚 , 𝒙 ≠ 𝒚 = 0, x=y.

Show that [^𝜹𝒇

𝜹𝒙]

(𝟎,𝟎) and [^𝜹𝒇

𝜹𝒙]

(𝟎,𝟎) exists.

Solution: We know by definition [^𝛿𝑓

𝛿𝑥]

(0,0) = lim

ℎ→0

𝑓(ℎ,0)−𝑓(0,0)

ℎ (if this limit exists).

= lim

ℎ→0 ℎ³−0

ℎ = 0, 𝑠𝑖𝑛𝑐𝑒 ℎ → 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 ℎ ≠ 0.

Similarly, [^𝛿𝑓

𝛿𝑦]

(0,0)

= lim

ℎ→0

𝑓(0,𝑘)−𝑓(0,0)

𝑘 (if this limit exists).

= lim

ℎ→0

−𝑘³−0

𝑘 = 0, 𝑠𝑖𝑛𝑐𝑒 𝑘 → 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑘 ≠ 0.

So both [^𝜹𝒇

𝜹𝒙]

(𝟎,𝟎) and [^𝜹𝒇

𝜹𝒙]

(𝟎,𝟎) exists and equal to 0.

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables We define the second order partial derivative with at (a,b) by

𝒇_𝒙𝒚(𝒂, 𝒃) = [^𝜹𝟐𝒇

𝜹𝒙𝜹𝒚]

(𝒂,𝒃) = [^𝜹

𝜹𝒙(^𝜹𝒇

𝜹𝒚)]

(𝒂,𝒃)

= 𝐥𝐢𝐦

𝒉→𝟎

𝒇_𝒚(𝒂+𝒉,𝒃)−𝒇_𝒚(𝒂,𝒃)

𝒉 (if this limit exists).

Similarly we define,

𝒇_𝒚𝒙(𝒂, 𝒃) = [ 𝜹^𝟐𝒇 𝜹𝒚𝜹𝒙]

(𝒂,𝒃)

= [𝜹 𝜹𝒚(𝜹𝒇

𝜹𝒙)]

(𝒂,𝒃)

= 𝐥𝐢𝐦

𝒌→𝟎

𝒇_𝒙(𝒂,𝒃+𝒌)−𝒇_𝒙(𝒂,𝒃)

𝒌 (if this limit exists).

Problem: Let 𝒇(𝒙, 𝒚) = ^{𝒙𝒚(𝒙𝟐− 𝒚𝟐)}

𝒙^𝟐+ 𝒚^𝟐 , 𝒘𝒉𝒆𝒓𝒆 (𝒙, 𝒚) ≠ (𝟎, 𝟎) = 0, (x,y) = (0,0).

Prove that ^𝜹𝟐𝒇

𝜹𝒙𝜹𝒚 ≠ ^𝜹𝟐𝒇

𝜹𝒚𝜹𝒙 at the origin.

Solution:

We have [^𝜹𝟐𝒇

𝜹𝒙𝜹𝒚]

(𝟎,𝟎) = 𝐥𝐢𝐦

𝒉→𝟎

𝒇_𝒚(𝒉,𝟎)−𝒇_𝒚(𝟎,𝟎)

𝒉

Now 𝑓_𝑦(ℎ, 0) =

lim

𝑘→0

𝑓(ℎ,𝑘)−𝑓(ℎ,0)

𝑘

= lim

𝑘→0

hk(h2− k2)

(h2+ k2) − ^{h.0(h2− 0)}

(h2+ 0)

k

= lim

𝑘→0

hk(h²− k²) k(h²+ k²)

=lim

𝑘→0

h(h²− k²) (h²+ k²)

=

^ℎ3

ℎ²

= ℎ.

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables Now 𝑓_𝑦(0,0) =

lim

𝑘→0

𝑓(0,𝑘)−𝑓(0,0)

𝑘

= lim

𝑘→0

^k.0(0−k2)

(k2+ 0) − 0

k

= 0.

[𝛿²𝑓 𝛿𝑥𝛿𝑦]

(0,0)

= lim

ℎ→0

𝑓

_𝑦⁽

ℎ, 0

⁾

− 𝑓

_𝑦

(0,0)

ℎ = lim

ℎ→0

ℎ − 0 ℎ = 1.

𝐴𝑙𝑠𝑜 [𝛿²𝑓 𝛿𝑦𝛿𝑥]

(0,0)

= lim

ℎ→0

𝑓

_𝑥⁽

0, 𝑘

⁾

− 𝑓

_𝑥

(0,0)

𝑘 .

Now 𝑓_𝑥(0, 𝑘) =

lim

ℎ→0

𝑓(ℎ,𝑘)−𝑓(0,𝑘)

ℎ

= lim

ℎ→0

hk(h2− k2) (h2+ k2) − 0

h

=lim

ℎ→0

k(h²− k²)

(h²+ k²)

=

^−𝑘3

𝑘²

= −𝑘.

And 𝑓_𝑥(0,0) =

lim

ℎ→0

𝑓(ℎ,0)−𝑓(0,0)

ℎ

= lim

ℎ→0 0−0

h

= 0.

[𝛿²𝑓 𝛿𝑦𝛿𝑥]

(0,0)

= lim

ℎ→0

𝑓

_𝑥⁽

0, 𝑘

⁾

− 𝑓

_𝑥

(0,0)

𝑘 = lim

ℎ→0

−𝑘 − 0

𝑘 = −1.

Therefore

^𝛿2𝑓

𝛿𝑥𝛿𝑦

≠

^𝛿2𝑓

𝛿𝑦𝛿𝑥

.

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables Differentiability: Let (a,b) be a point of the domain D⊂R² of the function f(x,y) and (h,k)be a pair of real numbers so small that (a+h,b+k) is in D. Then f(x,y) is said to be differentiable at (a,b) if we can express

𝒇(𝒂 + 𝒉, 𝒃 + 𝒌) − 𝒇(𝒂, 𝒃) = 𝑨𝒉 + 𝑩𝒌 + 𝒉𝝓(𝒉, 𝒌) + 𝒌𝝍(𝒉, 𝒌).

Where A and B are two real numbers independent of h, k; Ф and ψ are two functions of h, k such that 𝝓 → 𝟎, 𝝍 → 𝟎 as (𝒉, 𝒌) → (𝟎, 𝟎).

Alternative definition of differentiability: Let (a,b) be a point of the domain D⊂R² of the function f(x,y) and (h,k)be a pair of real numbers so small that (a+h,b+k) is in D.

Then f(x,y) is said to be differentiable at (a,b) if we can express

𝒇(𝒂 + 𝒉, 𝒃 + 𝒌) − 𝒇(𝒂, 𝒃) = 𝒉𝒇_𝒙(𝒂, 𝒃) + 𝒌𝒇_𝒚(𝒂, 𝒃) + 𝒉𝝓(𝒉, 𝒌) + 𝒌𝝍(𝒉, 𝒌).

Where 𝜑 and ψ are two functions of h, k such that 𝝓 → 𝟎, 𝝍 → 𝟎 as (𝒉, 𝒌) → (𝟎, 𝟎).

Theorem: If a function f(x,y) is differentiable at a point (a,b) of fixed domain. Then 1. Both fx(a,b) and fy(a,b) exist.

2. f(x,y) is continuous at (a,b).

Proof: Refer to textbook.

Problem: Discuss the differentiability of f(x,y) where 𝒇(𝒙, 𝒚) = ^𝒙𝒚

√𝒙^𝟐+ 𝒚^𝟐, 𝒘𝒉𝒆𝒓𝒆 (𝒙, 𝒚) ≠ (𝟎, 𝟎) = 0, (x,y) = (0,0).

Solution: To discuss the differentiability of f(x,y) at (0,0) let us find the partial derivatives fx(0,0) and fy(0,0).

𝑓_𝑥(0,0) =

lim

ℎ→0

𝑓(ℎ,0)−𝑓(0,0)

ℎ

(if this limit exists)

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables

= lim

ℎ→0 0−0

h

= 0.

Again 𝑓_𝑦(0,0) =

lim

𝑘→0

𝑓(0,𝑘)−𝑓(0,0)

𝑘

(if this limit exists)

= lim

𝑘→0 0 − 0

k

= 0.

Then the function f(x,y) is said to be differentiable at (0,0) if

𝑓(0 + ℎ, 0 + 𝑘) − 𝑓(0,0) = ℎ𝑓_𝑥(0,0) + 𝑘𝑓_𝑦(0,0) + ℎ𝜙(ℎ, 𝑘) + 𝑘𝜓(ℎ, 𝑘), where 𝜙 → 0, 𝜓 → 0 as (ℎ, 𝑘) → (0,0).

i.e.,

𝑓(ℎ, 𝑘) = ℎ. 0 + 𝑘. 0 + ℎ𝜙(ℎ, 𝑘) + 𝑘𝜓(ℎ, 𝑘),

⟹ 𝑓(ℎ, 𝑘) = ℎ𝜙(ℎ, 𝑘) + 𝑘𝜓(ℎ, 𝑘), ⟹ ^ℎ𝑘

√ℎ²+ 𝑘² = ℎ𝜙(ℎ, 𝑘) + 𝑘𝜓(ℎ, 𝑘)

⟹_√2ℎ^ℎ2₂= ℎ𝜙(ℎ, ℎ) + ℎ𝜓(ℎ, ℎ), setting h=k;

⟹ ¹

√2= 𝜙(ℎ, ℎ) + 𝜓(ℎ, ℎ);

where 𝜙 → 0, 𝜓 → 0 as (ℎ, 𝑘) → (0,0).

If f(x,y) is differentiable at (0,0) then passing through the limit as ℎ ⟶ 0 we will have consistent result.

But passing through the limit as ℎ ⟶ 0 we have ^𝟏

√𝟐= 𝟎, which is inconsistent.

So the function f(x,y) is not differentiable at (0,0).

C9T(SEM IV), Unit-I, Continuity and differentiability of fn of several variables Mean Value Theorem for a function of two variables:

Let f(x,y) be a function of two variables defined in some certain neighborhood N of (a,b) such that f(x,y) posses partial derivative ^𝝏𝒇

𝝏𝒙 in N of (a,b) and ^𝝏𝒇

𝝏𝒚 exists at the point (a,b). Then for any point (a+h,b+k) in N

𝒇(𝒂 + 𝒉, 𝒃 + 𝒌) − 𝒇(𝒂, 𝒃) = 𝒉𝒇_𝒙(𝒂 + 𝜽𝒉, 𝒃 + 𝒌) + 𝒌[𝒇_𝒚(𝒂, 𝒃) + 𝜼], Where 𝟎 < 𝜃 < 1 and 𝜼 is a function of k such that 𝜼 → 𝟎 𝒂𝒔 𝒌 → 𝟎; 𝒉, 𝒌 ∈ 𝑹.

Sufficient condition for differentiability:

If (a,b) be a point in the domain of definition of a function f of two independent variables x and y such that one of the partial derivatives 𝒇_𝒙 or 𝒇_𝒚 exist and the other is continuous at (a,b).

Suppose

1. ^𝝏𝒇

𝝏𝒙 exists at (a,b) and 2. ^𝝏𝒇

𝝏𝒚 is continuous at (a,b).

Then f(x,y) is differentiable at (a,b).

Proof: Refer to textbook.

C9T(SEM IV), Unit-I, Homogeneous fn, Chain rule Homogeneous Function:

Let f(x,y) be a function of two variables x and y defined in some domain D⊂R² . Then f(x,y) is said to be homogeneous function of degree n if we can write it

i) 𝒇(𝒙, 𝒚) = 𝒚^𝒏𝝋(𝒙/𝒚), ii) 𝒇(𝒙, 𝒚) = 𝒙^𝒏𝝋(𝒚/𝒙),

iii) 𝒇(𝒕𝒙, 𝒕𝒚) = 𝒕^𝒏𝒇(𝒙, 𝒚), ∀𝒕 ∈ 𝑹 𝒂𝒏𝒅 𝒕 ≠ 𝟎.

Example: Find the degree of homogeneity of the function 𝒇(𝒙, 𝒚) = 𝒙^𝟐+ 𝟔𝒙𝒚 + 𝟐𝒚^𝟐. Solution: Setting x=tx and y=ty we have

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡²𝑥²+ 6𝑡²𝑥𝑦 + 2𝑡²𝑦² = 𝑡²(𝑥²+ 6𝑥𝑦 + 2𝑦²) = 𝑡²𝑓(𝑥, 𝑦)

Therefore, f(x,y) is a homogeneous function of degree 2.

Euler’s theorem of Homogeneous function of two variables:

Let f(x,y) be a homogeneous function of two independent variables x and y defined in some domain D⊂R² of degree n. Then

𝒙^𝝏𝒇

𝝏𝒙+ 𝒚^𝝏𝒇

𝝏𝒚= 𝒏𝒇(𝒙, 𝒚).

Proof: Refer to textbook.

Converse Part of Euler’s theorem of Homogeneous function of two variables:

Let f(x,y) be a function of two independent variables x and y defined in some domain D⊂R² such that f(x,y) posses continuous first order partial derivatives in Dand

𝒙^𝝏𝒇

𝝏𝒙+ 𝒚^𝝏𝒇

𝝏𝒚= 𝒏𝒇(𝒙, 𝒚), ∀(𝒙, 𝒚)𝒊𝒏 𝑫;

then f(x,y) will be a homogeneous function of two independent variables x and y of degree n.

Proof: Refer to textbook.

C9T(SEM IV), Unit-I, Homogeneous fn, Chain rule

Problems: Let f be a homogeneous function of x and y of degree n. Then 𝒙^𝟐𝝏^𝟐𝒇

𝝏𝒙^𝟐+ 𝟐𝒙𝒚 𝝏^𝟐𝒇

𝝏𝒙𝝏𝒚+ 𝒚^𝟐𝝏^𝟐𝒇

𝝏𝒚^𝟐 = 𝒏(𝒏 − 𝟏)𝒇.

Solution: Since f be a homogeneous function of x and y of degree n then by Euler’s theorem we can write

𝑥^𝜕𝑓_𝜕𝑥+ 𝑦^𝜕𝑓

𝜕𝑦= 𝑛𝑓(𝑥, 𝑦) (1) Differentiating partially (1) with respect to x we get

𝑥^𝜕_𝜕𝑥^2𝑓₂+^𝜕𝑓

𝜕𝑥+ 𝑦 ^𝜕2𝑓

𝜕𝑥𝜕𝑦= ^𝜕𝑓

𝜕𝑥 𝑜𝑟, 𝑥𝜕²𝑓

𝜕𝑥²+ 𝑦 𝜕²𝑓

𝜕𝑥𝜕𝑦= (𝑛 − 1)𝜕𝑓

𝜕𝑥 (2) Differentiating partially (1) with respect to y we get

𝑥_{𝜕𝑦𝜕𝑥}^𝜕2𝑓 +^𝜕𝑓

𝜕𝑦+ 𝑦^𝜕2𝑓

𝜕𝑦² = 𝑛^𝜕𝑓

𝜕𝑦 𝑥_{𝜕𝑦𝜕𝑥}^𝜕2𝑓 + 𝑦^𝜕2𝑓

𝜕𝑦²= (𝑛 − 1)^𝜕𝑓

𝜕𝑦 (3)

Multiplying (2) by x and (3) by y and adding we get

𝑥^{2 𝜕}_𝜕𝑥^2𝑓₂+ 2𝑥𝑦 ^𝜕2𝑓

𝜕𝑥𝜕𝑦+ 𝑦^{2 𝜕2𝑓}

𝜕𝑦² = (𝑛 − 1)𝑥^𝜕𝑓

𝜕𝑥+ (𝑛 − 1)𝑦^𝜕𝑓

𝜕𝑦

= (n-1) { 𝑥^𝜕𝑓_𝜕𝑥 + 𝑦^𝜕𝑓

𝜕𝑦} =n(n-1)f by (1) Therefore

𝑥^{2 𝜕2𝑓}

𝜕𝑥²+ 2𝑥𝑦 ^𝜕2𝑓

𝜕𝑥𝜕𝑦+ 𝑦^{2 𝜕2𝑓}

𝜕𝑦²= 𝑛(𝑛 − 1)𝑓.(Proved)

C9T(SEM IV), Unit-I, Homogeneous fn, Chain rule Problem: If 𝒖 = 𝐭𝐚𝐧^{−𝟏 𝒙𝟑+𝒚𝟑}

𝒙−𝒚 then show that 𝒙^𝟐𝝏^𝟐𝒖

𝝏𝒙^𝟐+ 𝟐𝒙𝒚 𝝏^𝟐𝒖

𝝏𝒙𝝏𝒚+ 𝒚^𝟐𝝏^𝟐𝒖

𝝏𝒚^𝟐 = (𝟏 − 𝟒 𝐬𝐢𝐧^𝟐𝒖) 𝐬𝐢𝐧 𝟐𝒖.

Solution: Given that 𝒖 = 𝐭𝐚𝐧^{−𝟏 𝒙𝟑+𝒚𝟑}

𝒙−𝒚 . Therefore 𝐭𝐚𝐧 𝒖 = ^𝒙_𝒙−𝒚^{𝟑+𝒚𝟑}= 𝒗 (𝒔𝒂𝒚).

Then v is a homogeneous function of x and y of degree 2. Then by Euler’s theorem we can write

𝑥^𝜕𝑣_𝜕𝑥+ 𝑦^𝜕𝑣

𝜕𝑦= 2𝑣 (1) ⟹ 𝑥_𝜕𝑥^𝜕 (tan 𝑢) + 𝑦 ^𝜕

𝜕𝑦(tan 𝑢) = 2 tan 𝑢 ⟹ 𝑥 sec²𝑢^𝜕𝑢

𝜕𝑥 + 𝑦 sec²𝑢^𝜕𝑢

𝜕𝑦= 2 tan 𝑢 ⟹ 𝑥^𝜕𝑢

𝜕𝑥 + 𝑦^𝜕𝑢

𝜕𝑦= sin 2𝑢 (2) Differentiating partially (2) with respect to x we get

𝑥𝜕²𝑢

𝜕𝑥²+𝜕𝑢

𝜕𝑥+ 𝑦 𝜕²𝑢

𝜕𝑥𝜕𝑦= 2 cos 2𝑢𝜕𝑢

𝜕𝑥 Or,

𝑥𝜕²𝑢

𝜕𝑥²+ 𝑦 𝜕²𝑢

𝜕𝑥𝜕𝑦= (2 cos 2𝑢 − 1)𝜕𝑢

𝜕𝑥 (3) Differentiating partially (2) with respect to y we get

𝑥_{𝜕𝑦𝜕𝑥}^𝜕2𝑢 +^𝜕𝑢

𝜕𝑦+ 𝑦^𝜕2𝑢

𝜕𝑦²= 2 cos 2𝑢^𝜕𝑢

𝜕𝑦 Or, 𝑥_{𝜕𝑦𝜕𝑥}^𝜕2𝑢 + 𝑦^𝜕2𝑢

𝜕𝑦² = (2 cos 2𝑢 − 1)^𝜕𝑢

𝜕𝑦 (4) Multiplying (3) by x and (4) by y and adding we get

𝑥^{2 𝜕}_𝜕𝑥^2𝑢₂+ 2𝑥𝑦 ^𝜕2𝑢

𝜕𝑥𝜕𝑦+ 𝑦^{2 𝜕2𝑢}

𝜕𝑦² = (2 cos 2𝑢 − 1){ 𝑥^𝜕𝑢_𝜕𝑥+ 𝑦^𝜕𝑢

𝜕𝑦}

= (2 cos 2𝑢 − 1) sin 2𝑢 (From (1)) = (1 − 4 sin²𝑢) sin 2𝑢. (proved)

C9T(SEM IV), Unit-I, Exercise

1. Show that the limit exists at the origin but the repeated limit does not for the function

𝑓(𝑥, 𝑦) = 𝑥 sin1

𝑦+ 𝑦 sin1

𝑥, 𝑥𝑦 ≠ 0 = 0, xy=0.

2. Let 𝑓(𝑥, 𝑦) =^𝑥_𝑥³₂^{− 𝑦}_{+ 𝑦}³₂, 𝑤ℎ𝑒𝑟𝑒 (𝑥, 𝑦) ≠ (0,0) = 0, (x,y) = (0,0).

Show that f is not differentiable at (0,0) although f is continuous at (0,0) and fx ,fy both exist at (0,0).

3. Is the function 𝑓(𝑥, 𝑦) = √|𝑥𝑦| differentiable at (0,0). Justify.

4. If 𝑢 = sin^{−1 𝑥+𝑦}

√𝑥−√𝑦 then show that

𝑥

^{2 𝜕2𝑢}

𝜕𝑥²

+ 2𝑥𝑦

^𝜕2𝑢

𝜕𝑥𝜕𝑦

+ 𝑦

^{2 𝜕2𝑢}

𝜕𝑦²

= −

sin 𝑢 cos 2𝑢 4 cos³𝑢 .

5. Let H be a homogeneous function of x and y of degree n and 𝑢(𝑥, 𝑦) = (𝑥²+ 𝑦²)^−𝑛/2.

Then show that

_𝜕𝑥^𝜕 (𝐻^𝜕𝑢

𝜕𝑥) + ^𝜕

𝜕𝑦(𝐻^𝜕𝑢

𝜕𝑦) = 0.