Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

Given v (x, y) = x² - xy + \(\frac14\) y + 7, x, y ∈ R
dv = 2x dx - (x dy + ydx) \(\frac14\) (2y)dy + 0
dv = (2x-y)dx + (-x + \(\frac12\)y) dy

First let us find w_x, w_y, and w_z
Now w_x = 2xy + z², w_y = 2yz +x² and w_z = 2zx + y².
Thus,by (15), the differential is
dw = (2xy + z² )dx + (2yz + x² )dy+ (2zx + y² )dz.

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 G(x, y) = e^x+3y log (x² + y²)
\(\frac { \partial G }{ \partial x } ={ e }^{ x+3y }\frac { 1 }{ { x }^{ 2 }+y^{ 2 } } (2x)+log({ x }^{ 2 }+{ y }^{ 2 }){ e }^{ x+3y }\)
= \(\frac { 2x.{ e }^{ x+3y } }{ { x }^{ 2 }+{ y }^{ 2 } } +{ e }^{ x+3y }log({ x }^{ 2 }+{ y }^{ 2 })\)
= \({ e }^{ x+3y }\)
\(\left[ \left( \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 } } \right) +log({ x }^{ 2 }+{ y }^{ 2 }) \right] \)
\(\therefore { \left( \frac { \partial G }{ \partial x } \right) }_{ (-1,1) }={ e }^{ -1+3 }\left[ \frac { -2 }{ 2 } +log(2) \right] \)
\(={ e }^{ 2 }(log2-1)\)
\(\frac { \partial G }{ \partial x } ={ e }^{ x+3y }\frac { 1 }{ { x }^{ 2 }+y^{ 2 } } (2y)+log{ (x }^{ 2 }+y^{ 2 }){ e }^{ x+3y }(3)\)
\(={ e }^{ x+3y }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 } } +3log({ x }^{ 2 }+{ y }^{ 2 }) \right) \)
\({ \left( \frac { \partial G }{ \partial x } \right) }_{ (-1,1) }={ e }^{ -1+3 }\left( \frac { -2 }{ 2 } +3log2 \right) \)
\(={ e }^{ 2 }(1+log{ 2 }^{ 3 })\)
= \({ e }^{ 2 }(1+log8)\)

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)
\(\left( \frac { \partial h }{ \partial z } \right) =0+x(2z)=2xz\)
\(\therefore { \left( \frac { \partial h }{ \partial x } \right) }_{ \left( 2,\frac { \pi }{ 4 } ,1 \right) }=2(2)(1)=4\)

Given g(x, y) = 3x² + y² + 5x + 2
\(\frac { { \partial }g }{ { \partial x } } \) = 6x + 0+ 5 = 6x + 5
\({ \left( \frac { { \partial }g }{ { \partial x } } \right) }_{ (1,-2) }\) = 6(1) + 5 = 11
\(\frac { { \partial }g }{ { \partial y } } \) = 0 + 2y + 0 + 0 = 2y
\(\therefore { \left( \frac { { \partial }g }{ { \partial y } } \right) }_{ (1,-2) }\) = 2(-2) = -4

Given f(x, y) = 3x² - 2xy + y² + 5x + 2, (2,-5)
\(\frac { { \partial }f}{ { \partial x } } \) = 3x² - 2y(1) + 0 + 2
\({ \left( \frac { { \partial }f}{ { \partial x } } \right) }_{ (2,5) }={ 3(2 }^{ 2 })-2-(-5)+5\)
= 12 + 10 + 5 =27
\(\frac { { \partial }f }{ { \partial y } } \) = 0 - 2x(1) + 2y + 0 + 0
\({ \left( \frac { { \partial }f }{ { \partial x } } \right) }_{ (2,-5) }\) = 2(-5)-(2)(2)
= -10 - 4 = -14

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.

Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g is continuous at every point of R²

Let f(x) = ^{\(x^\frac13\)}, xo = 27, ∆x = -1
∴ \(\sqrt [ 3 ]{ 26 } =f(27)+{ f }^{ ' }(27)\)
\(f(27)={ (27) }^{ \frac { 1 }{ 3 } }={ (2^{ 3 }) }^{ \frac { 1 }{ 3 } }={ 3 }^{ 1 }=3\)
\({ f }^{ ' }(x)=\frac { 1 }{ 3 } x^{ \frac { 1 }{ 3 } -1 }=\frac { 1 }{ 3 } { x }^{ \frac { 2 }{ 3 } }=\frac { 1 }{ { 3x }^{ \frac { 2 }{ 3 } } } \)
∴ \({ f }^{ ' }(27)=\frac { 1 }{ { 3(27)x }^{ \frac { 2 }{ 3 } } } =\frac { 1 }{ { { { 3(3 }^{ 3 } }) }^{ \frac { 2 }{ 3 } } } =\frac { 1 }{ 3({ 3 }^{ 2 }) } =\frac { 1 }{ 27 } \)
∴ (1) becomes
\(\sqrt [ 3 ]{ 26 } =3+\frac { 1 }{ 27 } (-1)\)
= 3 - \(\frac{1}{27}\)(-1)
= 3 - \(\frac{1}{27}\) = 3 - 0.037
\(\sqrt [ 3 ]{ 26 } \) = 2.963

Let f(x) = ^{\(x^\frac14 \)}, x₀ 16, Δx = -1
∴ \(\sqrt [ 4 ]{ 15 } \) = f(16) +f'(16) (-1) ... (1)
\(f(16)={ (16) }^{ \frac { 1 }{ 4 } }={ (2^{ 4 }) }^{ \frac { 1 }{ 4 } }={ 2 }^{ 1 }=2\)
\({ f }^{ ' }(16)={ \frac { 1 }{ 4 } x }^{ \frac { 1 }{ 4 } -1 }={ \frac { 1 }{ 4 } x }^{ -\frac { 3 }{ 4 } }=\frac { 2 }{ { 4x }^{ \frac { 3 }{ 4 } } } \)
\({ f }^{ ' }(125)=\frac { 1 }{ { 6({ 2 }^{ 4 })x }^{ \frac { 3 }{ 4 } } } =\frac { 1 }{ { { 4(2 }^{ 3 } }) } =\frac { 1 }{ 4(8) } =\frac { 1 }{ 32 } \)
∴ becomes
\(\sqrt [ 4 ]{ 15 } =2+\frac { 1 }{ 32 } (-1)\)
\(=2-\frac { 1 }{ 32 } =2-0.0312\)
\(\sqrt [ 4 ]{ 15 } =1.968 \)

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

Let us calculate the limit of f as (x, y)→(0, 0) along the line y = x.
Then \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) f (x, y) =0 ; because along the line y = x when f (0,0) = 1 ≠ 0;
Hence f cannot be continuous at (0, 0).

We compute
F(\(\lambda \)x, \(\lambda \)y) = \(\frac { { \left( { \lambda x } \right) }^{ 2 }+5\left( \lambda x \right) \left( \lambda y \right) -10(\lambda y)^2 }{ 3\lambda x+7\lambda y } =\frac { { \lambda }^{ 2 } }{ \lambda } \left( \frac { { x }^{ 2 }+5xy-10{ y }^{ 2 } }{ 3x+7y } \right) =\lambda F(x,y)\)
for all \(\lambda \) ∈ R. So F is a homogeneous function of degree 1.

Given \(U(x,y,z)=xy+sin\left( \frac { { y }^{ 2 }-2{ x }^{ 2 } }{ xy } \right) \)
\(u(\lambda x,\lambda y,\lambda z)=\lambda x\lambda y+sin\left( \frac { { \lambda }^{ 2 }{ y }^{ 2 }-2{ \lambda }^{ 2 }{ z }^{ 2 } }{ \lambda x\lambda y } \right) \)
\(={ \lambda }^{ 2 }xy+sin\left( \frac { { y }^{ 2 }-2{ x }^{ 2 } }{ xy } \right) \)
≠ λp. u (x, y, z)
There is no common λ
\(\therefore\) It is not homogeneous.

Given \(g(x,y,z)=\frac { \sqrt { { 3 }x^{ 2 }+5{ y }^{ 2 }+{ z }^{ 2 } } }{ 4x+7y } \)
\(g(\lambda x,\lambda y,\lambda z)=\frac { \sqrt { 3{ \lambda }^{ 2 }{ \lambda }^{ 2 }+5{ \lambda }^{ 2 }{ y }^{ 2 }+{ \lambda }^{ 2 }{ z }^{ 2 } } }{ 4zx+7\lambda y } \)
\(=\frac{\not x \sqrt{3 x^{2}+5 y^{2}+z^{2}}}{\not \lambda(4 x+7 y)}\)
= λ⁰g(x, y, z)
Thus g is homogeneous with degree 0.

\(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
Given \(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
\(h(\lambda x,\lambda y)=\frac { 6{ \lambda }^{ 2 }{ x }^{ 2 }{ \lambda }^{ 3 }{ y }^{ 3 }-\pi { \lambda }^{ 5 }{ y }^{ 5 }+9{ \lambda }^{ 4 }{ x }^{ 4 }\lambda y }{ 2020{ \lambda }^{ 2 }{ x }^{ 2 }+2019{ \lambda }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { { \lambda }^{ 5 }(6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y) }{ { \lambda }^{ 2 }(2020{ x }^{ 2 }+2019{ y }^{ 2 }) } \)
\( =\lambda^{3} \mathrm{~h}(\mathrm{x}, \mathrm{y}) \)
Thus f is homogeneous with degree 3.

Given f(x, y) = x²y + 6x³ + 7
f(λx, λy) = λ²x²λy + 6λ³x³ + 7
= λ³ x² y + 6λ³ x³ + 7
≠ λf(x, y)
There is no common λ in this equation.
It is not homogeneous

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

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}=\left| \frac { { y }^{ 2 }-xy }{ \sqrt { x } -\sqrt { y } } -0 \right| \)
= \(\left| \frac { { y }^{ 2 }-xy }{ \sqrt { x } -\sqrt { y } } \right| =\frac { \left| y \right| \left| y-x \right| }{ \left| \sqrt { x } -\sqrt { y } \right| } \)
= \(\frac { \left| y \right| \left| \sqrt { x } +\sqrt { y } \right| |\sqrt { y } -\sqrt { x } | }{ \left| -\sqrt { y } -\sqrt { x } \right| } \)
\(=\frac{|y||\sqrt{x}+\sqrt{y}| \sqrt{y}-\sqrt{\not x} \mid}{|\sqrt{y}-\sqrt{\not x}|}\)
= \(|y||\sqrt { x } +\sqrt { y } |\)
\(\therefore \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}=\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}|y||\sqrt { x } +\sqrt { y } |=0\)

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
\(dy=\frac { (3-4x)[3{ (1-2x) }^{ 2 }(-2)]-({ 1-2x) }^{ 3 }(-4) }{ { (3-4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-3(3-4x)+2(1-2x)] }{ (3-{ 4x) }^{ 2 } } dx\)
= \(\frac { { 2(1-2x) }^{ 2 }[-9+12x+2-4x] }{ { (3-4x) }^{ 2 } } dx\)
\(dy=\frac { { 2(1-2x) }^{ 2 }[8x-7] }{ { (3-4x) }^{ 2 } } dx\)

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

Given w(x, y) = 6x³ -3xy + 2y²x = e^x; cos(s)
\(\frac { \partial w }{ \partial x } ={ 18x }^{ 2 }-3y;\frac { \partial w }{ \partial y } =3x+4y\)
= 18 (e^4s) - 3 cos (s);
\(\frac { \partial w }{ \partial y } =-3{ e }^{ s }+4cos(s)\)
\(\frac { dx }{ ds } ={ e }^{ s };\frac { dy }{ ds } =-sin(s)\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial w }{ \partial x } .\frac { dx }{ ds } +\frac { \partial w }{ \partial y } .\frac { dy }{ ds } \)
= [18 e^s - 3 cos (s)]e^s + (-3e^s + 4 cos (s)). (- sin (s))
\(\therefore \frac { dw }{ ds } \) = 18e^s - 3e^s cos (s) + 3e^3s (sin s) - 4 sin s cos s
Now, \({ \left( \frac { dw }{ ds } \right) }_{ s=0 }\) =18(1)-3(1)(1)+0-0
= 18 - 3 = 15

Given w (x, y, z) = x² +y² +y², 
x = e^t, y = e^t sin t, z = e^t cos t
\(\frac { \partial u }{ \partial x } \) = 2x; \(\frac { \partial u }{ \partial y} \) = 2y; \(\frac { \partial u }{ \partial z} \) = 2z
\(\frac { \partial u }{ \partial x } \) = 2e^t
\(\frac { \partial u }{ \partial y} \) = 2e^t sin t
\(\frac { \partial u }{ \partial z} \) = 2e^t cos t
\(\frac{dx}{dt}\) = e^t
\(\frac{dy}{dt}\) = e^t cas t + sin t e^t
⇒ \(\frac{dz}{dt}\) = et (- sin t ) + cos t et
By chain rule
\(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { dw }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } \) = 2e^t(e^t) + 2e^t sin t (e^t cos t +sin t e^t) - e^t sin t + 2e^t cas t (e^t cos t - e^t sin t )

= e^2t [2 + 2] = 4e^2t

Given u(x, y, z) = xy²z³, x = sin t, y = cos t, z = 1+ e^2t
\(\frac { \partial u }{ \partial x } ={ y }^{ 2 }{ z }^{ 3 };\frac { \partial u }{ \partial y } ={ 2xyz }^{ 3 }\)
\(\frac { \partial u }{ \partial z } =3{ x }y^{ 2 }{ z }^{ 2 }\)
\(\frac { \partial u }{ \partial x } ={ cos }^{ 2 }t+{ (1+{ e }^{ 2t }) }^{ 3 };\)
\(\frac { \partial u }{ \partial y } =2sin \ t \ cos \ t{ (1+{ e }^{ 2t }) }^{ 3 }\);
\(\frac { \partial u }{ \partial z } =3 \ sin \ t \ { cos }^{ 2 }t{ (1+{ e }^{ 2t }) }^{ 3 }\)
\(\frac { dx }{ dt } =cos \ t;\frac { dy }{ dt } =-sin \ t\)
\(\frac { dz }{ dt } ={ 2e }^{ 2t }\)
By chain rule,
\(\frac { du }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } +\frac { \partial u }{ \partial z } .\frac { dz }{ dt } \)
= cos² t (1 + et)³ (cos t) + 2 sin t cos t ( 1+ e²t)³ (- sin t) + 3 sin t cos²t (1+ e²t)² (2e²t)
= (1 + e²t)2 [cos³ t(1 + e²t) - 2 sin² t cos t (1+ e^2t) + 6 sin (cos² t e^2t]
\(\frac{du}{dt}\) = (1 + e^2t)² [cos³t (1 + e^2t) - sin t sin 2t (1 e^2t) + 6 e^2t sin t cos²t 
[∵ sin 2t = 2 sin t cos t]

Given u (x, y) = x²y + 3xy⁴, x = e^t, y = sin t
\(\frac { du }{ dt } =\frac { \partial u }{ \partial x } .\frac { \partial x }{ \partial t } +\frac { \partial w }{ \partial y } .\frac { \partial y }{ \partial t } \) ...(1)
x = e^t
⇒ \(\frac { dx }{ dt } ={ e }^{ t }\)
y = sin t
⇒ \(\frac { dy }{ dt } =cos \ t\)
\(\frac { \partial u }{ \partial x } =2xy+3{ y }^{ 4 }\);
\(\frac { \partial u }{ \partial y } ={ x }^{ 2 }+12{ xy }^{ 3 }\)
⇒ \(\frac { \partial u }{ \partial x } =2{ e }^{ t }sint+3{ sin }^{ 4 }t\)
\(\frac { \partial u }{ \partial y } ={ e }^{ 2t }+12{ e }^{ t }{ sin }^{ 3 }t\)
Substituting in (1) we get,
\(\frac{du}{dt}\) = (2e^t sin t + 3 sin⁴t)e^t + (e²t + 12e^t sin³ t) cos t
\(\frac{du}{dt}\) = 2e^2tsin t + 3 e^t sin⁴ t + e^2t cos t + 12e^t sin⁴ t cos t
= e^t[2e^t sin t + 3 sin⁴ t + cos t + 12 sin³ t cos t]
\({ \left( \frac { du }{ dt } \right) }_{ t=0 }\) = e⁰[2(1) (0) +3(0) + 1+ 12 (0)]
= 1 [1] = 1
∴ \({ \left( \frac { du }{ dt } \right) } \) = 1

Given v(x, y, z) = x³ + y³ + z³ + xyz³
\(\frac { \partial v }{ \partial z } \) = 0 + 0 +3z² + 3xy = 3z² + 3xy
\(\frac { \partial v }{ \partial y } \) = 0 + 3y² + 0 + 3xz = 3y² + 3xz
Now, \(\frac { { \partial }^{ 2 }v }{ \partial y\partial z } =\frac { { \partial } }{ \partial { y } } \left( \frac { \partial v }{ \partial z } \right) \) = 0 + 3x = 3x ...(1)
\(\frac { { \partial }^{ 2 }v }{ \partial z\partial y } =\frac { { \partial } }{ \partial { z } } \left( \frac { \partial v }{ \partial y } \right) \) = 0 + 3x = 3x ..... (2)
From (1) and (2),
\(\frac { { \partial }^{ 2 }v }{ \partial y\partial z } =\frac { { \partial }^{ 2 }v }{ \partial z\partial y } \)

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.

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

Note that the function f takes value 1 on the x, y-axes and 0 everywhere else on R². So let us calculate
\(\frac { \partial f }{ \partial x } (0,0)\) = \(\underset { h\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0+h,0 \right) -f(0,0) }{ h } =\underset { h\longrightarrow 0 }{ lim } \frac { 1-1 }{ h } =0;\)
\(\frac { \partial f }{ \partial y } (0,0)\) = \(\underset { k\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0,0+k \right) -f(0,0) }{ k } =\underset { k\longrightarrow 0 }{ lim } \frac { 1-1 }{ k } =0\)
This completes (i).