g(x, y) = \(\frac { { x }^{ 2 }y }{ { x }^{ 4 }+{ y }^{ 2 } } \) Also y = kx²
∴ g(x, y) = \(\frac { { x }^{ 2 }.k.{ x }^{ 2 } }{ { x }^{ 4 }+{ k }^{ 2 }{ x }^{ 4 } } =\frac { { x }^{ 4 }k }{ { x }^{ 4 }(1+{ k }^{ 2 }) } =\frac { k }{ 1+{ k }^{ 2 } } \)
∴ \(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\) \(\frac { { x }^{ 2 }y }{ { x }^{ 4 }+{ y }^{ 2 } } =\frac { k }{ 1+{ k }^{ 2 } } \)
along every parabola y = kx², k ∈ R \ {0}.

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.

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

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

Given y = _ex^2_-5x+7 cos (x² - 1)
Taking differentials,
dy = (_ex^2_-5x+7 (-sin (x² - 1)(2x)) +cos (x² - 1) _ex^2_-5x+7 (2x - 5) dx
= _ex^2_-5x+7 [(2x - 5)cos (x² - 1) - 2x sin (x² - 1)]dx

Given = (3 + sin(2x)) ^2/3 
Taking differentilas,
dy =  \(\frac23\)(3 + sin(2x)) ^2/3-1 (cos 2x) (2)dx
dy = \(\frac { 4 }{ 3 } .\frac { cos2x }{ { (3+sin2x) }^{ \frac { 1 }{ 3 } } } dx\)

Given y = \(​​y=\frac { { \left( 1-2x \right) }^{ 3 } }{ 3-4x } \)
Taking differentials
\(dy=\frac { (3-4x)[3{ (1-2x) }^{ 2 }(-2)]-({ 1-2x) }^{ 3 }(-4) }{ { (3-4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-3(3-4x)+2(1-2x)] }{ (3-{ 4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-9+12x+2-4x] }{ { (3-4x) }^{ 2 } } dx\)
\(dy=\frac { { 2(1-2x) }^{ 2 }[8x-7] }{ { (3-4x) }^{ 2 } } dx\)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

\(\frac { \partial z }{ \partial x } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } (y)+{ tan }^{ -1 }(xy)\)
\(\frac { \partial z }{ \partial y } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)(x)
\(\frac { \partial z }{ \partial z } =\frac { xy }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \) + tan^-1 (xy)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial x } =\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } +{ tan }^{ -1 }({ t }^{ 2 }{ se }^{ 2 })\)
\(\frac { \partial z }{ \partial y } =\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\frac { dx }{ dt } =2t;\frac { dy }{ dt } ={ s.e }^{ t }\)
Also, \(\frac { dx }{ ds } =0;\frac { dy }{ ds } ={ e }^{ t }\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial z }{ \partial y } .\frac { dy }{ ds } \)
= \(\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } (0)+\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } { (e }^{ t })=\frac { { e }^{ t }{ t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\therefore { \left( \frac { \partial z }{ \partial s } \right) }_{ s=t=1 }=\frac { e(1) }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
By chain rule
= \(\frac { dz }{ dt } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
\(\therefore { \left( \frac { \partial z }{ \partial t } \right) }_{ (s=t=1) }=\frac { e }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
= \(\frac { 2e+e }{ 1+{ e }^{ 2 } } =\frac { 3e }{ 1+{ e }^{ 2 } } \)+ 2 tan^-1(e)

Given g (x, y) = x² + 3xy - 7y + cos(5x)
g_x = 2x + 3y - 0 - 5 sin 5x
= 2x + 3y - 5 sin 5x
g_y = 0 + 3x (1) - 7 + 0
= 3x - 7
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\) = 3
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })\)
= 2(1) + 0 - 5(5) cos (5x)
= 2 - 25 cos (5x)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\) 0
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })\)
= 0 + 3(1) - 0 = 3

g(x, y) = log (5x + 3y)
\({ g }_{ x }=\frac { 1 }{ 5x+3y } (5)=\frac { 5 }{ 5x+3y } \)
\({ g }_{ y }=\frac { 1 }{ 5x+3y } (3)=\frac { 3 }{ 5x+3y } \)
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\)
= 3(-1)(5x + 3y)^-2 (5)
\(=\frac { -15 }{ { (5x+3y) }^{ 2 } } \)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })=\frac { -5 }{ ({ 5x+3y) }^{ 2 } } (5)\)
\(=\frac { -25 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\frac { -3 }{ ({ 5x+3y) }^{ 2 } } (3)\)
\(=\frac { -9 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })=\frac { -5 }{ 5x+3y } (3)\)
\(=\frac { -15 }{ ({ 5x+3y) }^{ 2 } } \)

g(x, y) = xe^y + 3x²y
g_x = e^y + 6xy
gy = xey + 3x²
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })={ e }^{ y }+6x\)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ xy })=0+6y=6y\)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })={ xe }^{ y }\)
\({ g }_{ xy }=\frac { \partial }{ \partial y } ({ g }_{ x })\) = e^y + 6

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