w = xy + z
\(\frac { \partial w }{ \partial x } =y;\frac { \partial w }{ \partial y } =x;\frac { \partial w }{ \partial z } =1\)
⇒ \(\frac { \partial w }{ \partial x } \) = sin t; \(\frac { \partial w }{ \partial y } \) = cos t; \(\frac { \partial w }{ \partial z } \) = 1
\(\frac { dx }{ dt } \) = - sin t; \(\frac { dy}{ dt } \) = cos t; \(\frac { dz }{ dt } \) = 1
∴  \(\frac { dw }{ dt } \) = \(\frac { \partial w }{ \partial x } .\frac { dx }{ dt } +\frac { \partial w }{ \partial y } .​​\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
= sin t(-sin t) + cos t(cost) + 1 (1) 
= - sin² t + cos² t + 1
= cos² t + 1 - sin² t
= cos² t + cos² t  [∵ 1- sin² t = cos² t]
\(\frac { dw }{ dt } \) = 2 cos² t

Let x_o = 3 and dx = 0.02
f(x_o) = f(3) = 3 (3²) + 5 (3) + 3
= 27 + 15 + 3 = 45
f'(x) = 6x + 5
f'(x) = f'(3) = 6 (3) + 5 = 23
∴ f(3. 02) = f(x_o) +f'(x_o) dx
= 45 + 23 (.02)
= 45 + 0.46 = 45.46

Given g(x, y) = \(\frac { { e }^{ y }sinx }{ x } \) for x ≠ 0 and g(0, 0) = 1
g(0, 0) = 1
The function g is defined for all (x, y) ∈ R²
To check if g has a limit L at (0,0) and if L= g(0,0) = 1
Consider \(\left| g(x,y)-g(0,y) \right| =\left| \frac { { e }^{ y }sinx }{ x } -0 \right| \)
= \(\left| \frac { { e }^{ y }sinx }{ x } \right| =\left| \frac { \left| { e }^{ y } \right| \left| sinx \right| }{ x } \right| =\left| { e }^{ x } \right| \left| \frac { sinx }{ x } \right| =1\)
\(\left[ \because (x,y)\longrightarrow (0,0)\Rightarrow \left| { e }^{ y } \right| =1and\left| \frac { sinx }{ x } \right| =1 \right] \)
\(\because \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\frac { { e }^{ y }sinx }{ x } =1=g(0,0)\) Which proves that is continuous at (0, 0)
∴ g(x, y) is continuous at (0, 0)

Given a = edge of the circle
= 10 cm and da
= 0.2 cm
Volume of cube = a³
Approximate amount of cubic centimeters of paint is used to paint this cube = 3a² da
= 3(10²)(0.2)
= 300 \(\left( \frac { 2 }{ 10 } \right) \) = 60 cm³
Exact amount of paint used = f(x + ∆x) - f(x)
= f(10.2) - f(10)
= 10.2³ -10³
= 1061.208 - 1000
= 61.208 cm²

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Actual value = 12.5 cm,
Approximate value = 12.65 cm
Area of the circular plate = πr²
Relative error = \(\frac{160.225π-156.25π}{160.0225π}\)
= \(\frac{3.7725π}{160.0225π}\)= 0.024 cm

Given \(g(x)=\sqrt { { x }^{ 2 }+9 } ,{ x }_{ 0 }=-4\)
\(g(x)=\sqrt { { (-4) }^{ 2 }+9 } =\sqrt { 16+9 } =5\)
\({ g }^{ ' }(x)=\frac { 1 }{ 2 } ({ { x }^{ 2 }+9 })^{ -\frac { 1 }{ 2 } }(2x)=\frac { x }{ \sqrt { { x }^{ 2 }+9 } } \)
\(\therefore { g }^{ ' }({ x }_{ 0 })=\frac { -4 }{ \sqrt { { (-4) }^{ 2 }+9 } } =\frac { -4 }{ 5 } \)
∴ L(x) = g(x_o) + g'(x₀)(x - x_o)
= \(5-\frac { 4 }{ 5 } (x+4)=\frac { 25-4x-16 }{ 5 } \)
L(x) = \(\frac { 9-4x }{ 5 } \)

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 r² = x² +y² + z²
log r² = log (x² + y² + z²)
⇒ 2 log r = log (x² + y² + z²)
∴ 2V = log (x² + y² + z²) [∵ V = log r ]
⇒ V = \(\frac12\) log (x² + y² + z²)
\(\frac { \partial V }{ \partial x } =\frac { 1 }{ 2 } \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =\frac { x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } =\frac { ({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 })(1)-x(2x) }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
III^ty \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ y }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
∴ \(\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { y }^{ 2 }+{ z }^{ 2 }-{ x }^{ 2 }+{ z }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } \)
= \(\frac { { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } } \)
= \(\frac { 1 }{ { r }^{ 2 } } \)
Hence proved.

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
\(\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } { \left( \frac { \partial f }{ \partial x } \right) }=4\)
\(\frac { { \partial }^{ 2 }f }{ { \partial y }^{ 2 } } =\frac { \partial }{ \partial y } { \left( \frac { \partial f }{ \partial y } \right) }=6\)

We need to show that u satisfies the Laplace’s equation in R². Observe that u_x(x, y) = e^-2y(-2)sin(2x) and hence u_xx (x, y) = e^-2y(-2)(2) cos(2x).
Similarly, u_y( x y) = e^-2y (-2)cos(2x) and u_yy (x, y) = (-2)(-2)e^-2ycos(2x)
Thus, u_xx + u_yy = -4e^-2y cos(2x) + 4e^-2y cos(2x) = 0.

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

Let y = f(x) = \(\sqrt x\)
Let x_o = 25, dx = 25.2 - 25 = 0.2
y = \(\sqrt x\)
dy = \(\frac{1}{2\sqrt{x}}\) dx
dy = \(\frac{1}{2\sqrt{x}}\) (0.2) = 0.02
∴\(\sqrt{25.2}\) = f(x₀) + f'(x₀) dx
= \(\sqrt{25}\) + 0.02
= 5 + 0.02 = 5.02