\(L(z)={ 2 }^{ 1/4 }+\frac { 1 }{ 4 } \left( { 2 }^{ -3/4 } \right) \left( z-2 \right) \)

Given v (x, y) = x² - xy + \(\frac14\) y + 7, x, y ∈ R
dv = 2x dx - (x dy + ydx) \(\frac14\) (2y)dy + 0
dv = (2x-y)dx + (-x + \(\frac12\)y) dy

Given w (x, y) = x³ - 3xy+ 2.0, x, y ∊ R
w_x = \(\frac { \partial w }{ \partial x } \) = 3x² - 3y + 0 = 3x² - 3y 
w_x = at (1, -1)= 3 (1p - 3- (-1) = 3 + 3 = 6
wy = \(\frac { \partial w }{ \partial y } \) = 0 -3x+4y = -3x+4y
w_y (1, -1) = -3 (1)+4(-1)= -3 - 4 = -7.
w (x_o, y_o) = w (1, -1)
= 1³- 3(1) (-1) + 2 (-1)²
= 1+3+2 = 6
Linear approximation is given by
L(x,y) = w(x_o,y_o) + \((\frac { \partial w }{ \partial x })_{(x_o,y_o) }\)(x-x₀) + \((\frac { \partial w }{ \partial xy})_{(x_o,y_o) }\) (y-y_o)
= 6 + 6 (x - 1) + -7 (y + 1)
= 6 + 6x - 6 - 7y - 7
= 6x - 7y - 7

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.

\({ h }({ x }_{ o })=\frac { x }{ 1+1 } =\frac { 1 }{ 2 } \)
\({ h }^{ ' }(x)=\frac { (x+1)(1)-x(1) }{ { (x+1) }^{ 2 } } \)
\(\frac { x+1-x }{ { (x+1) }^{ 2 } } =\frac { 1 }{ ({ x+1) }^{ 2 } } \)
\({ h }^{ ' }({ x }_{ o })=\frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ 4 } \)
∴ L(x) = h(x_o) + h'(x₀)(x - x_o)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } (x-1)=\frac { 2+x-1 }{ 4 } =\frac { x+1 }{ 4 } \)
∴ L(x) = \(\frac { x+1 }{ 4 } \)

Let f(x) = \(f(x)={ x }^{ \frac { 2 }{ 3 } },{ x }_{ 0 }=125,\triangle x=-2\)
∴ (123)^\(\frac23\) = f(125) +1'(125) (-2) ... (1)
\(f(125)={ (125) }^{ \frac { 2 }{ 3 } }={ { (5 }^{ 3 }) }^{ \frac { 2 }{ 3 } }\) = 5² = 25
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } \)

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,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =3f\)

Let r be the radius and A be the area of the plate
Given \(\frac { \triangle r }{ r } \times 100=2\)
when r = 10
\(\frac { \triangle r }{ r } \times 100=2\)
⇒ \(\triangle r=\frac { 2r }{ 100 } =\frac { 2\times 10 }{ 100 } =\frac { 2 }{ 10 } \)
\(\therefore dr=\frac { 2 }{ 10 } \)
A= ㅠr²
dA = 2πr(dr) = 2π(10)\(\left( \frac { 2 }{ 10 } \right) \)
= 4πcm²