W(x, y, z) = xy + yz + zx, x =u -v, y = uv, z = u + v; y = uv; z = u
\(\frac { \partial W }{ \partial x } \) = y + z; \(\frac { \partial W }{ \partial y } \) = x + z
∴ \(\frac { \partial W }{ \partial x } \) = uv + u +  v;
\(\frac { \partial W }{ \partial y} \) = u - v + u + v;
\(\frac { \partial W }{ \partial z} \) = uv +  u - v
\(\frac { dx }{ du } =1;\frac { dy }{ du } =v;\frac { dz }{ du } =1\)
\(\frac { dx }{ dv } =1;\frac { dy }{ dv } =v;\frac { dz }{ dv} =1\)
By chain rule
\(\frac { \partial W }{ \partial u } =\frac { \partial w }{ \partial x } .\frac { dx }{ du } +\frac { \partial w }{ \partial y } .\frac { dy }{ du } +\frac { \partial w }{ \partial z } .\frac { dz }{ du } \)
= (uv +u +v) (1) +2u (v) + (uv +u - v)(1)

\(\frac { \partial W }{ \partial u } \) = 4uv + 2u = 12u (2v + 1)
\({ \left( \frac { \partial W }{ \partial u } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = 2 x \(\frac12\) (2+ 1) = 1(2+ 1) = 3
= (uv + u + v) (-1) + (2u) (u) + (uv + u - v)(1)

= 2u² - 2v = 2 (u² - v)
∴  \({ \left( \frac { \partial W }{ \partial v } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = \(2\left( \frac { 1 }{ 4 } -1 \right) =2\left( -\frac { 3 }{ 4 } \right) =-\frac { 3 }{ 2 } \)

Given w (x, y) = xy + sin (xy)
\(\frac { \partial w }{ \partial x } \) = y (1) + (cos (xy) [y (1)]
= y + y cos (xy)
\(\frac { \partial w }{ \partial y } \) = x (1) + cos (xy) (x)
= x + x cos (xy)
\(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) = \(\frac { { \partial } }{ \partial { y } } \left( \frac { \partial w }{ \partial x } \right) \)
= 1 + y (- sin (xy)) (x) + cos (xy)
= 1 -xy sin (xy) + cos (xy) ... (1)
\(\frac { { \partial } }{ \partial { x } } \left( \frac { \partial w }{ \partial y } \right) \)
= 1 + x (- sin (xy)) (y) + cos (xy)
= 1 - xy sin (xy) + cos (xy) ... (2)
∴ From (1) and (2),
\(\frac { { \partial }^{ 2 }w }{ \partial x\partial y } =\frac { { \partial }^{ 2 }w }{ \partial x\partial y } \)

Given V(x, y) = e^x(x cos y - y sin y)
\(\frac { \partial V }{ \partial x } \) = e^x (cos y) +(x cos y - y sin y)e^x
= e^x (cos y + x cos y - y sin y)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) = e^x(0 + cos y - 0) + (cos y +x cos y.- y sin y)e^x
= e^x(2 cos y + x cos y - y sin y) ... (1)
\(\frac { \partial V }{ \partial y } \) = e^x(-x sin y- y cos y- sin y)

\(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x (-x cos y - (-y sin y + cos y) - cos y)
= e^x(- x cos y + y sin y - cos y - cos y)
= e^x (- x cos y + y sin y - 2 cos y) ... (2)
(1)+(2)➝ 
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } \) + \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } \) = e^x(2 cos y + x cos y - y sin y - x cos y + y sin y - 2 cos y]
= e^x (0) = 0
Hence proved

Given w(x, y, z) = \(\frac { 1 }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } }\)
= (x² + y² + z²) -^\(\frac12\)
\(\frac { \partial w }{ \partial x } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2x)\)
= (-x² + y² + z²) -^\(\frac12\)
\(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial w }{ \partial x } \right) \)
= -[x\(\left( \frac { -3 }{ 2 } \right) \)( x² + y² +z²)^\(-\frac32\)
\((\not 2 x)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}\)
= (x² + y² + z²)^-\(\frac52\) [-3x² + x² +y + z²]
= - (x² + y² + z²)^-\(\frac52\) [y² + z² - 2x²] ....(1)
\(\frac { \partial w }{ \partial y } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2y)\)
= -y(x² +y² + z²)-^\(\frac32\) 
\(\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } =\frac { \partial }{ \partial y } \left( \frac { \partial w }{ \partial y } \right) \)
\(=-\left[y\left(\frac{-3}{\not 2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{\frac{5}{2}}(\not 2 y)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}(1)\right]\)
= -(x² + y² + z²)-^{\(\frac52 \)} 
= -(x² + y² + z²)-^{\(\frac52 \)} [3y² + x² + y² + z²]
= -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² + 2z²] ....(2)
Now \(\frac { \partial w }{ \partial z } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2z)\)
= -z(x² + y² + z²)-^{\(\frac32 \) 
∴} \(\frac { { \partial }^{ 2 }w }{ \partial { z }^{ 2 } } \) = -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² - 2z²] ...(3)
(1)+(2)+(3)⟶
^∴ \(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } \) ⁼ -(x² + y² + z²)-^{\(\frac52 \)} [y² + z² - 2x² + x² + z² - 2y + x²+ y-2z²]
= -(x² + y² + z²)-^{\(\frac52 \)}(0) = 0
Hence proved 

First we shall calculate \(\frac { \partial f }{ \partial x } \) (x, y). Note that f is a sum of two functions and so
\(\frac { \partial f }{ \partial x } =\frac { \partial }{ \partial x } sin({ xy }^{ 2 })+\frac { \partial }{ \partial x } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial x } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial x } \left( { x }^{ 3 }+5y \right) \)
= cos(xy² ) y² + \({ e }^{ { x }^{ 3 }+5y }\) 3x²
Similarly,
\(\frac { { \partial }^{ }f }{ { \partial y\ } } =\frac { \partial }{ \partial y } sin(x{ y }^{ 2 })+\frac { \partial }{ \partial y } \left( { e }^{ { x }^{ 3 }+5y } \right) \)
\(=cos(x{ y }^{ 2 })\frac { \partial }{ \partial y } (x{ y }^{ 2 })+{ e }^{ { x }^{ 3 }+5y }\frac { \partial }{ \partial y } \left( { x }^{ 3 }+5y \right) \)
= cos(xy²)2xy + 5\({ e }^{ { x }^{ 3 }+5y }\)
Next we consider,
\(\frac { { \partial }^{ 2 }f }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 })+3{ x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
\(=\frac { \partial }{ \partial y } ({ y }^{ 2 }cos(x{ y }^{ 2 }))+\frac { \partial }{ \partial y } ({ 3x }^{ 2 }{ e }^{ { x }^{ 3 }+5y })\)
= 2y cos(xy²)+y²(-sin(xy²)2xy) + 3x² \({ e }^{ { x }^{ 3 }+5y }\) 5
= 2y cos(xy²)+2xy³ sin(xy²)+15 x² \({ e }^{ { x }^{ 3 }+5y }\)
Finally,
\(\frac { { \partial }^{ 2 }f }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial f }{ \partial x } \right) =\frac { \partial }{ \partial y } (cos(x{ y }^{ 2 })2xy+5{ e }^{ { x }^{ 3 }+5y })\)
= -sin(xy²)y²2xy+cos(xy²)2y+5\({ e }^{ { x }^{ 3 }+5y }\) 3x²
= 2y cos(xy²)- 2xy³ sin(xy²)+15x²\({ e }^{ { x }^{ 3 }+5y }\)
Note that we have first used sum rule, then in the next step we have used chain rule. In the third step, product rule is used. Also, we see that f_xy = f_yx Is it a coincidence? or is it always true? Actually, there are functions for which f_xy ≠ f_yz at some points. The following theorem gives conditions under which f_xy = f_yz.

First we shall calculate \(\frac { \partial F }{ \partial x } \)(x, y) then we evaluate it at (−1, 3) As we have already observed we find the derivative with respect to x holding y as a constant. That is,
\(\frac { \partial f }{ \partial x } (x,y)=\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial x } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial x } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial x } +\frac { \partial (7) }{ \partial x } \)
= 3x² y + y² +0
= 3x² y + y² .
so, \(\frac { \partial F }{ \partial x } \) ( -1, 3) = 3( -1)² 3 + 3² = 18.
Next similarly we find partial derivative with respect to y.
\(\frac { \partial F }{ \partial y } \) (x, y) = \(\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial y } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial y } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial y } +\frac { \partial (7) }{ \partial y } \)
= x³ + 2yx + 0
= x³ + 2yx.
Hence we have \(\frac { \partial F }{ \partial y } \) (-2, 1) = (-2)³ + 2(1)( -2) = -12.
Note that in the above example \(\frac { \partial F }{ \partial x } \) (x, y) = 3x² y + y² which is again a function of two variables. 
So, we can take the partial derivative of this function with respect to x or y.
For instance, if we take G(x, y) = 3x² y+y² then we find  \(\frac { \partial F }{ \partial x } \) = 6xy. Since G(x, y) = \(\frac { \partial F }{ \partial x } \), we have \(\frac { \partial G }{ \partial x } \)=\(\frac { \partial }{ \partial x } \)\(\left( \frac { \partial f }{ \partial x } \right) \) = 6xy.
We denote this as \(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } \) which is called the second order partial derivative of F with respect to x. 
Also, \(\frac { \partial F }{ \partial y } \) = 3x² + 2y. Since G (x, y) = \(\frac { \partial F }{ \partial x } \) we have  \(\frac { \partial G }{ \partial y } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \) = 3x² + 2y. 
We denote this as \(\frac { { \partial }^{ 2 }F }{ \partial y\partial x } \) which is called the mixed partial derivative of F with respect to x, y. 
Similarly we can also calculate \(\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \right) \) = 3x²+2y.
Also, if we differentiate \(\frac { \partial F }{ \partial y } \) partially with respect to y we obtain \(\frac { { \partial }^{ 2 }F }{ { \partial y }^{ 2 } } \) which is called the second order partial derivatives of F with respect to y.
So for any function F defined on any subset {(x, y) | a < x < b, c < y < d} ⊂ R² we have the following notation
\(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ xx' }\frac { { \partial }^{ 2 }F }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ xy }\)
\(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ yx' }\frac { { \partial }^{ 2 }F }{ { { \partial y }^{ 2 } } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ yy }\)
All the above are called second order partial derivatives of F. 
Similarly we can define higher order partial derivatives.
For example, \(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) and \(\frac { { \partial }^{ 2 }F }{ { \partial x\partial y\partial x } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) 
Next we shall see more examples on partial differentiation.

When x = 4, dx = 4.1-4 = 0.1

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
 

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

Given  W(x, y, z) = x² - xy + 3 sin z, x, y, z ∈ R
W(x₀, y₀, z₀) = W (2, -1, 0)
= 2²-2(-1)+3 sin 0
\(\frac { \partial W }{ \partial x } \) = 2x - y + 0 = 2x - y
\(\left( \frac { \partial W }{ \partial x } \right) _{(2, -1, 0)}\)= 2(2)-(-1) = 5
\(\frac { \partial W }{ \partial y } \) = 0-x + 0 = -x
\(\left( \frac { \partial W }{ \partial y } \right) _{(2, -1, 0)}\) = -2
\(\frac { \partial W }{ \partial z } \) = 0 - 0 + 3 cos z
 \(\left( \frac { \partial W }{ \partial z} \right) _{(2, -1, 0)}\)= 3 cos 0 = 3(1) = 3.
Linear approximation is given by
L(x, y, z) = w(x₀, y₀, z₀) + \(\left( \frac { \partial W }{ \partial x } \right) _{(x_0, y_0, z_0)}\) (x - x₀) + \(\left( \frac { \partial W }{ \partial y} \right) _{(x_0, y_0, z_0)}\) (y - y₀) + \(\left( \frac { \partial W }{ \partial z} \right) _{(x_0, y_0, z_0)}\)(z₀ - z₀)
∴ L(x,y,z) = 6 + 5(x - 2) - 2(y+1) + 3(z - 0)
= 6 + 5x - 10 - 2y - 2 + 3z
L (x, y, z) = 5x - 2y + 3z - 6

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 r = 10.5 cm
dr = 10.75 - 10.5 = 0.25
Area = πr²
Approximate change in area = π(2r) dr
= 2π(10.5)(0.25)
= 5.25 π cm²
Approximate percentage change in the area
= \(\frac{5.25 π}{π(10.5)(10.5)}\) x 100
= \(\frac{5.25}{110.25}\) x 100
= 0.0476 x 100 = 4.765%

V(t) = 30 + 12t² - t³, 0 ≤ t ≤ 8
Given t = 4, dt = 4\(\frac16\) - 4 = \(\frac16\)
Approximate change in voters = (24t -3t²)dt
= [24(4) - 3(4)²] = \(\frac16\)
= (96 - 48)\(\frac16\) = \(\frac{48}{6}\) = 8
Since the function is given in thousands approximate change in voters = 8000

Let x be the number
Let y = f(x) = \(x^\frac{1}{n}\)
Then log y = \(\frac1n\) log x
Taking differential on both sides we get,
\(\frac { 1 }{ y } dy=\frac { 1 }{ n } \times \frac { 1 }{ x } dx\)
i.e. \(\frac { \Delta y }{ y } \simeq \frac { dy }{ y } =\frac { 1 }{ n } .\frac { dx }{ x } \)
\(\therefore \frac { \Delta y }{ y } \times 100\simeq \frac { 1 }{ n } \left( \frac { dx }{ x } \times 100 \right) \)
\(\simeq \frac { 1 }{ n } \) times the percentage error in the number. Hence, percentage error in the nth root of a number is approximately \(\frac1n\) times the percentage error in the number

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 = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

Actual value = 12.5 cm,
Approximate value = 12.65 cm
Area of the circular plate = πr²
Percentage error = 0.024 x 100 = 2.4%
Volume of sphere = \(\frac43\)πr² 

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

Actual radius of the circular plate = 12.5 cm
Measured radius of the circular plate = 12.65
dr = 12.65-12.5
= 0.15
\( \mathrm{A} =\pi \mathrm{r}^{2} \\ \mathrm{dA} =2 \pi \mathrm{rdv} \)
Change in Area
A(12.65)-A(12.5) = dA
\( =2 \pi \times 12.5 \times 0.15 \)
\(=3.75 \pi \)
Absolute error = 3.7725\(\pi\) - 3.75\(\pi\)
:0.0225\(\pi\) cm²

Given R (x, y) = 80 x + 90 y + 0.04xy - 0.05 x² + 0.05 y² and
(x,y) = 8x + 6y + 2000
Profit function P (x,y) = Revenue - cost
P (x,y) = R (x,y) - C (x,y)
= -80 x + 90 y + 0.04xy - 0.05 x² - 0.05y² - 8x - 6y - 2000
P (x, y) = 72x + 84y + 0.04 xy - 0.05 x² - 0.05y² - 2000
(ii) \(\frac { { \partial P } }{ \partial { x } } \) = 72 + 0 + 0.04y - 0.05(2x) - 0 - 0
= 72 + 0.04y- 0.1x
∴ \(\frac { { \partial P } }{ \partial { x } } \) (1200, (1800)
= 72+ 0.04 (1800) - 0.1(1200)
= 72 + 72 - 120 = 24 .......(1)
\(\frac { \partial v }{ \partial y} \) = 0 + 84+ 0.4x-0-0.5(2y) - 0
= 84 + 0.04x - 0.1y
= 84 + 0.04 (1200) - 0.1(1800)
∴ \(\frac { \partial v }{ \partial y} \)_(1200,1800) = 84 + 48 - 180 = - 48 .......(2)
From (1) and (2), keeping y constant and 4 increasing x then increases profit.

y = 52 \(\sqrt { x } \), 0, ≤ x ≤ 9
Give x = 1, dx = 1.1-1 = 0.1
Approximate number of words learned
= \(52.\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }\) = \(\frac { 26 }{ \sqrt { x } } \) dx
= \(\frac { 26 }{ \sqrt { 1 } } \) (0.1) = 2.6
≃ 3 words
(ii) Then approximate number of words learned
= \(\frac { 26 }{ \sqrt { x } } \)dx = \(\frac { 26 }{ \sqrt { 4} } \)(0.1) = \(\frac{26}{4}\)(0.1)
= 13 (0.1) = 1.3
≃ 1 word