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

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)  ...(2)
∴ From (1) and (2), f_xy = f_yz

First we calculate  \(\frac { { \partial }w }{ { \partial x } } (x,y)=\frac { { \partial }(xy) }{ { \partial x } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial x } \)
This gives  \(\frac { { \partial }^{ }w }{ { \partial x } } \) (x, y) = y + 0 and hence \(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) (x, y) = 1 On the other hand,
\(\frac { { \partial }w }{ { \partial y } } (x,y)=\frac { { \partial }(xy) }{ { \partial y } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial y } \)
\(=x+\frac { \left( { y }^{ 2 }+1 \right) { e }^{ y }-{ e }^{ y }2y }{ \left( { y }^{ 2 }+1 \right) } \)
Hence, \(\frac { { \partial }^{ 2 }w }{ { \partial x\partial y } } \) (x, y) = 1

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.

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 (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

Given h (x, y, z) = x sin (xy) + z²x
\(\frac { \partial h }{ \partial x } =x.cos(x,y).\frac { \partial }{ \partial x } (xy)+sin(xy)(1)+{ z }^{ 2 }(1)\)
= x cos (xy) (y)(1) + sin (xy) + z²
= xy cos (xy) + sin (xy) + z²
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }=2\left( \frac { \pi }{ 4 } \right) cos\left( 2\frac { \pi }{ 4 } \right) +sin\left( 2\frac { \pi }{ 4 } (1) \right) +{ 1 }^{ 2 }\)
\(=\frac { \pi }{ 2 } cos\left( \frac { \pi }{ 2 } \right) +sin\left( \frac { \pi }{ 2 } \right) +1\)
\(=\frac { \pi }{ 2 } (0)+1+1=2\)
\(\frac { \partial h }{ \partial y } =x.cos(xy).\frac { \partial h }{ \partial y } (xy)+0\)
= x cos (xy) x(1)
= x² cos (xy)
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }={ 2 }^{ 2 }cos\left( 2\frac { \pi }{ 4 } \right) \)
= \(4cos\left( \frac { \pi }{ 2 } \right) \)
= 4(0) = (0)
\(\left( \frac { \partial h }{ \partial z } \right) =0+x(2z)=2xz\)
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }=2(2)(1)=4\)

Let (a, b)  ∈ R² an arbitrary point we shall investigate continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a, b).
(i) f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) is defined
(ii) \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}=\frac { \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ x }^{ 2 }-{ y }^{ 2 } }{ \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ y }^{ 2 }+1 } \) = L exists
= \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) = L exists
(iii) Also, f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \)
∴ \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}\)= L = f(a, b)
Hence f satisfies all the three conditions since (a, b) is an arbitrary point on R², we conclude that f is continuous at every point of R².

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

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
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

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.

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}cos=\left( \frac { { e }^{ x }siny }{ y } \right) \) = \(cos\left( { e }^{ 0 }\frac { siny }{ y } \right) \)
= cos[(1)(1)] = cos (1) \(\left[ \because \begin{matrix} lim \\ y\rightarrow 0 \end{matrix}\frac { siny }{ y } =1 \right] \)

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

Note that g is differentiable and g'(x) = 2x + cos x
Thus dg = (2x + cos x)dx.