Given w(x, y, z) = \(\left( \frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
Let (x, y, z) = \(\frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
⇒ w = log f  ...(1)
⇒ e^w = f
f(λx, λy, λz) = \(\frac { { 5\lambda }^{ 3 }{ x }^{ 3 }{ \lambda }^{ 4 }{ y }^{ 4 }+7{ \lambda }^{ 2 }{ y }^{ 2 }\lambda x{ \lambda }^{ 4 }{ z }^{ 4 }-75{ \lambda }^{ 3 }{ y }^{ 3 }{ \lambda }^{ 4 }{ z }^{ 4 }{ }^{ } }{ { \lambda }^{ 2 }{ x }^{ 2 }+{ \lambda }^{ 2 }{ y }^{ 2 } } \)
= \(\frac { { \lambda }^{ 7 }(5{ x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75 }y^{ 3 }{ z }^{ 4 } }{ { \lambda }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) } ={ \lambda }^{ 5 }f(x,y,z)\)
∴ f(x, y, z) is a homogeneous function of degree 5.
∴ By Euler's theorem,
\(x.\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } +z\frac { \partial f }{ \partial z } =5.f\)
⇒ \(x.\frac { \partial }{ \partial x } ({ e }^{ w })+y.\frac { \partial }{ \partial y } ({ e }^{ w })+z.\frac { \partial }{ \partial z } ({ e }^{ w })=5.{ e }^{ w }\) [using (1)]
⇒ \(x.{ e }^{ w }\frac { \partial w }{ \partial x } +y.{ e }^{ w }\frac { \partial w }{ \partial y } +z.{ e }^{ w }\frac { \partial w }{ \partial z } ({ e }^{ w })=5{ e }^{ w }\)
⇒ \(x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } \)  [Divided by e^w]

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 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 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]

Here again we shall use the tree diagram to calculate \(\frac { \partial g }{ \partial r } ,\frac { \partial g }{ \partial s } \) 
Hence we find \(\frac { \partial g }{ \partial x } \) = 2x, \(\frac { \partial g }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial s } \) = -1, \(\frac { \partial y }{ \partial r } \) =2r, and \(\frac { \partial y }{ \partial s } \) = 2.
Now, \(\frac { \partial g }{ \partial r } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial r } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial r } \) = 2x(2) + 2(2r) 12r - 4s.
also,  \(\frac { \partial g }{ \partial s } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial s } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial s } \) = 2x(-1) +(2)2 = 2s - 4r + 4.

We shall follow the tree diagram to calculate
So first we need to find \(\frac { \partial g }{ \partial x } ,\frac { \partial g }{ \partial y } ,\frac { dx }{ dt } \) and \(\frac { dx }{ dt } \).
Now, \(\frac { \partial g }{ \partial x } \) = 2x-y + cos(x + y), \(\frac { \partial g }{ \partial x } \) = -x+cos(x + y), \(\frac { dx }{ dt } \) = 3e^3t and \(\frac { dx }{ dt } \) = 2t.
Thus, \(\frac { dg }{ dt } =\frac { \partial g }{ \partial x } \frac { dx }{ dt } +\frac { \partial g }{ \partial y } \frac { dy }{ dt } \)
= (2x − y + cos(x + y)) 3e^3t + (−x + cos(x + y))( 2t)
= (2e^3t - t² + cos(e^3t - t²))3e^3t +( -e^3t + cos(e^3t - t²))(2t ) 
= 6e^6t - 3t² e^3t +3e^3t cos(e^3t - t²) -2te^3t +2t cos(e^3t - t²) 
Also, some times our W(x, y) will be such that x = x(s, t) , and y = y(s, t) where s, t ∈ R. Then W can be considered as a function that depends on s and t. If x, y both have partial derivatives with respect to s, t and W has partial derivatives with respect to x and y, then we can calculate the partial derivatives of W with respect to s and t using the following theorem.

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

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 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.

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

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

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

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

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).

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 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.

Given r = 2 mm
dr = (2.1 - 2) = 0.1 mm
Area = πr²
Approximate area dA = 2πr dr
= 2π (2) (0.1)
= 4 π (0.1) = 0.4 π mm²

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