Let f(x) = tan x, x_o= 45, dx = 1
f(x_o) = f(45) = tan 45 = 1
f'(x) = sec² x dx
f'(x_o) = f'(45) = sec² 45 (1)
\((\sqrt{2})^2\) =2 (0.01745)
= 0.03490
∴ tan 46° = f(x_o) +f'(x_o) dx
=- 1 + 0.03490 = 1.03490

Given g(x, y) = \(\frac { { x }^{ 2 }y }{ { x }^{ 4 }+{ y }^{ 2 } } \)
Given y = mx
∴ g(x, y) = \(\frac { { x }^{ 2 }.mx }{ { x }^{ 4 }+{ m }^{ 2 }{ x }^{ 2 } } \)
= \(\frac { m{ x }^{ 3 } }{ { x }^{ 2 }({ x }^{ 2 }+{ m }^{ 2 }) } =\frac { mx }{ { x }^{ 2 }+{ m }^{ 2 } } \)
Now, \(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}g(x,y)=\frac { m(0) }{ { 0 }^{ 2 }+{ m }^{ 2 } } =\frac { 0 }{ { m }^{ 2 } } =0\) for all values of m
∴ \(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\) g(x, y) = 0 along every line y = mx, m ∈ R.

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

Given u(x, y) = x² + 3xy + y²
u(x_o, y_o) = u(2,1)
= 2² + 3(2)(1) + 1²
= 4 + 6 + 1 = 11
\(\frac { \partial u }{ \partial x } \) = 2x+ 3y
\({ \left( \frac { \partial u }{ \partial x } \right) }_{ (2,1) }\)= 2 + 3 = 5
\(\frac { \partial u }{ \partial y } \) = 3x+ 2y
\({ \left( \frac { \partial u }{ \partial y } \right) }_{ (2,1) }\) = 6 + 2 = 8
Linear approximation 
L(x,y) = U(x_o, y_o) + \({ \left( \frac { \partial u }{ \partial x } \right) }_{ ({ x }_{ 0 },{ y }_{ 0 }) }\) (x - x_o) + \({ \left( \frac { \partial u }{ \partial y} \right) }_{ ({ x }_{ 0 }{ ,y }_{ 0 }) }\)(y - y_o)
L (x,y) = 11 + 5 (x - 2) + 8 (y - 1)
= 11 + 5x - 10 + 8y - 8
L(x,y) = 5x + 8y - 7

Given f(x, y) = 2x³ - 11x²y + 3y³
f(tx, ty) = 2t³ x3 - 11 t² x²ty + 3t³y³
= t³(2x³ - 11x²y + 3y³)
= t³. f(x,y)
∴ f (x, y) is a homogeneous function of degree 3.
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =3f\)

Given = (3 + sin(2x)) ^2/3 
Taking differentilas,
dy =  \(\frac23\)(3 + sin(2x)) ^2/3-1 (cos 2x) (2)dx
dy = \(\frac { 4 }{ 3 } .\frac { cos2x }{ { (3+sin2x) }^{ \frac { 1 }{ 3 } } } dx\)

Given f(x, y) = 2x² + 3y² - 2xy
\(\frac { \partial f }{ \partial x } \) = 4x - 8y
\({ \left( \frac { \partial f }{ \partial x } \right) }_{ (2,3) }\) = 4(2) - 8(3)
= 8 - 24 = -16
\(\frac { \partial f }{ \partial y } \) = 6y-8x
\({ \left( \frac { \partial f }{ \partial y } \right) }_{ (2,3) }\) = 6(3)- 8(2)
= 18-16 = 2
\(\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } { \left( \frac { \partial f }{ \partial x } \right) }=4\)
\(\frac { { \partial }^{ 2 }f }{ { \partial y }^{ 2 } } =\frac { \partial }{ \partial y } { \left( \frac { \partial f }{ \partial y } \right) }=6\)

Given g (x, y) = x log \(\left( \frac { y }{ x } \right) \)
g(λx, λy) = λx log \(\left( \frac { \lambda y }{ \lambda x } \right) \)
= λx log \(\left( \frac { \lambda y }{ \lambda x } \right) \)
= λ x log \(\left( \frac { y }{ x } \right) \)
= λ¹ g (x, y)
∴ g (x, y) is a homogeneous function of degree 1.
To verify \(x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } =1g\) [Euler's theorem]
∴ \(\frac { \partial g }{ \partial x } \) = \(x.\frac { 1 }{ \frac { y }{ x } } \left( -\frac { y }{ { x }^{ 2 } } \right) +log\left( \frac { y }{ x } \right) \)(1)
= \(-\frac { { x }^{ 2 } }{ y } \left( \frac { y }{ { x }^{ 2 } } \right) log\left( \frac { y }{ x } \right) \)
= - 1 + log \(\left( \frac { y }{ x } \right) \)
\(\frac { \partial g }{ \partial y } =x.\frac { 1 }{ \left( \frac { y }{ x } \right) } \left( \frac { 1 }{ x } \right) =\frac { 1 }{ \frac { y }{ x } } =\frac { x }{ y } \)
Consider \(x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } \)
= \(x\left( -1+log\frac { y }{ x } \right) +y\left( \frac { x }{ y } \right) \)

= x log \(\left( \frac { y }{ x } \right) \) = g
\(\therefore x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } \) = 1(g)
Hence, Euler's theorem is verified.

g(x, y) = log (5x + 3y)
\({ g }_{ x }=\frac { 1 }{ 5x+3y } (5)=\frac { 5 }{ 5x+3y } \)
\({ g }_{ y }=\frac { 1 }{ 5x+3y } (3)=\frac { 3 }{ 5x+3y } \)
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\)
= 3(-1)(5x + 3y)^-2 (5)
\(=\frac { -15 }{ { (5x+3y) }^{ 2 } } \)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })=\frac { -5 }{ ({ 5x+3y) }^{ 2 } } (5)\)
\(=\frac { -25 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\frac { -3 }{ ({ 5x+3y) }^{ 2 } } (3)\)
\(=\frac { -9 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })=\frac { -5 }{ 5x+3y } (3)\)
\(=\frac { -15 }{ ({ 5x+3y) }^{ 2 } } \)

Diameter = 30 cm
Radius = 15 cm
Circumference (c) = 2πr
\(\frac{dc}{dr}\) = 2π(3) = 6πcm
dc = 2πdr
\(\frac{6}{2π}\) cm = dr
\(\frac{3}{π}\) cm = dr
Approximate growth of the diameter
= 2dr = 2 \(\times\) \(\frac{3}{π}\) cm = \(\frac{6}{2π}\)cm
(ii) A = πr²
dA = π ²r dr
\(d \mathrm{~A}=\not \pi 2(15) \frac{3}{\not \pi} \mathrm{cm}^{2}\)
dA = 90 cm²
Area = πr² = π \(\times\)15 \(\times\) 15 cm²
 

(4) \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy } \) is a homogeneous function of  degree 0.