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

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
Percentage error = 0.024 x 100 = 2.4%
Volume of sphere = \(\frac43\)πr² 

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

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

Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

We know from (4), that L(x) = f(x₀) +f'(x₀)(x-x₀) We have x₀ = 3, \(\Delta \)x = 0.2 and hence f(3) = \(\sqrt { 1+3 } \) = 2. Also,
f′(x) = \(\frac { 1 }{ 2\sqrt { 1+x } } \) and hence f'(3) = \(\frac { 1 }{ 2\sqrt { 1+3 } } \) = \(\frac { 1 }{ 4 } \)
Thus, L(x) = 2 +\(\frac { 1 }{ 4 } \)(x-3) = \(\frac { x }{ 4 } \)+\(\frac { 5 }{ 4 } \) gives the required linear approximation.
Now, f(3.2) = \(\sqrt { 4.2 } \) ≈ L(3.2) = \(\frac { 3.2 }{ 4 } \)+\(\frac { 5 }{ 4 } \) = 2.050
Actually, if we use a calculator to calculate we get \(\sqrt { 4.2 } \) = 2.04939

2.0116 approx.

\( \frac { \partial w }{ \partial u } =\frac { 2u }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } ;\frac { \partial w }{ \partial v } =\frac { -2v }{ \sqrt { 1-\left( { u }^{ 2 }-{ v }^{ 2 } \right) } } \)

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(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 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.

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% 

\(\frac { { \vartheta }^{ 2 }u }{ \vartheta x\vartheta y } ={ e }^{ xy }\left[ 3xy+1+{ x }^{ 2 }{ y }^{ 2 } \right] \)

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 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

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

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(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