\(\frac{\partial u}{\partial x} =3 x^2+3 y^2 \)
\(\frac{\partial^2 u}{\partial y \partial x} =6 y \)

\(\frac{\partial u}{\partial x} =8 x+4 y+4 \)
\(\frac{\partial^2 u}{\partial y \partial x} =4 \)

\(\eta_d=-\frac{p}{x} \frac{d x}{d p}=\frac{-p}{\frac{1}{p}}\left(\frac{-1}{p^2}\right)=1\)

Given f(x, y) =3x²+4y³+6xy-x²y³+6
We know that f_y(x, y) 12y² + 6x - 3x²y²
Differentiating again partially w.r.t. 'x' we get,
f_xy(x,y) 0 + 6 - 3y² (2x) = 6 - 6xy²
\(\therefore\)f_xy(2, 1) = 6 - 6(2)(1)² = 6 - 12 = - 6

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating 'f' partially w.r.t. 'y' we get
f_y(x,y) = 0+ 12y²+6x(1)-x²(3y²)+0
=12y²+ 6x - 3x²y²
Differentiating again partially w.r.t. 'y' we get,
fy(x, y) =24y + 0 - 3x²(2y) = 24y - 6x²y
\(\therefore\)f_yy(1, 1)= 24(1) - 6(1²)(1) = 24 - 6 = 18

Given y=x-1/x
Differentiating w.r.t.'x' we get,
\({dy\over dx}=1+{1\over x^2}>0\) for all real values of x, except x=0
\(\therefore \) y is a strictly increasing function for all real values of x(x\(\neq\)0)

Given P = C(L)^α(K)^β
Differentiating partially w.r.t. 'K' we get
\({∂P\over ∂K}=C(L)^α\beta(K)^{β-1}=Cβ(L)^α(K)^{β-1}\)
\(K.{∂P\over ∂K}=Cβ.(L)^α(K)^{β-1}\)
=C(L)^αβ(K)^β-1+1=Cβ(L)^α(K)^β-1   ....(1)
Differentiating 'P' partially w.r.t. 'L' we get,
\({∂P\over ∂L}=C.(K)^β.α(L)^{α-1}=Cα(K)^{ \beta }(L)^{ { α-1 } }\)
\(L.{∂P\over ∂L}=Cα(K)^β.(L^1)(L)^{α-1}\)
=Cα(K)^β(L)^α-1+1=Cα(K)^β(L)^α .....(2)
Adding (1) and (2) we get
\({L{∂P\over ∂L}+K{∂P\over ∂K}}=Cβ(L)^\alpha (K)^β + C^\alpha (K)^β(L)^\alpha = C.(L)^\alpha(K)^β [β + \alpha]\)
= C.(L)^α (K)^β [1]     [∵ α + β = 1]
= C.(L)^α (K)^β
\(∴\ {L{∂P\over ∂L}+K{∂P\over ∂K}}=P\)
Hence Proved

Given \(u= e^{x/y} sin\left(x\over y\right)+e^{y/x}cos\left(y\over x\right)\)
\(u(tx,ty)=e^{tx\over ty}sin\left(tx\over ty\right)+e^{tx\over ty}cos\left(ty\over tx\right)\)
\(=e^{x\over y}sin\left(x\over y\right)+e^{y\over x}cos\left(y\over x\right)\)
= t⁰u(x,y)
∴ u is a homogenous function of degree 0.
∴ By Euler's theorem,
\(x.{∂u\over ∂x}+y.{∂u\over ∂y}=nu\)
\(⇒x\ .{∂u\over ∂x}+y.{∂u\over ∂y}=0u\)
\(⇒x.{∂u\over ∂x}+y.{∂u\over ∂y}=0\)
Hence proved

Given \(u=\sqrt{x^2+y^2}\)
Differentiating partially w.r.t. 'x' we get,
\({∂u\over ∂ x}={1\over 2}(x^2+y^2)^{1/2-1}(2x)=x(x^2+y^2)^{-1/2}={x\over \sqrt{x^2+y^2}}\)
\(∴\ x{∂u\over ∂x}={x^2\over\sqrt{x^2+y^2}}\)  ...(1)
\({∂u\over ∂y}={1\over2}(x^2+y^2)^{1/2-1}(2y)=y(x^2+y^2)^{-1/2}={y\over \sqrt{x^2+y^2}}\)
\(∴\ y{\partial u\over \partial x}={y^2\over \sqrt{x^2+y^2}}\)  ...(2)
(1)+(2) \(∴\ x{\partial u\over \partial x}+y{\partial u\over \partial y}={x^2+y^2\over \sqrt{x^2+y^2}}=\sqrt{x^2+y^2}\) ...(3)
Also \(u(tx,ty)=\sqrt{(tx)^2+(ty)^2}=\sqrt{t^2x^2+t^2y^2}=\sqrt{t^2(x^2+y^2)}=t{\sqrt{x^2+y^2}}=t^1u\)
\(\therefore\) u is a homogeneous function of degree 1.
\(\therefore x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial x } =nu\)
\(\Rightarrow x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial x } =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)       [From (3)]
Hence verified.     

Given R = 8x, \(C = 150000 + 60\left(x^2\over 900\right)\)
Profit = Revenue - Cost
\(P=8x- 150000- 60\left(x^2\over900\right)\)
Differentiating w.r.t. 'x' we get,
\({dP\over dx}=8-{120x\over 900}=8-{2x\over 15}\)
Condition for maximum is \(dp\over dx\)and \({d^2P\over dx^2}<0\)
\({dP\over dx}=0⇒8={2x\over 15}⇒{8\times15\over2}=x\)
⇒x=60
Also, \({d^2P\over dx^2}={-2\over 15}<0\)
∴ Profit is maximum when x = 60.
So, to get the maximum profit, we should sell 60 units.

Let y= 75-12x+6x²-x³
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=0-12+12x-3x^2=-3(x^2-4x+4)=-3(x-2)^2\)
\(\Rightarrow{dy\over dx}\le 0\) for all \(x \in (-\infty , \infty)\)
\(\therefore \) y decreases as x increases.

Given y=1+1/x
Differentiating w.r.t. 'x' we get,
\({dy\over dx}={-1\over x^2}<0\) for all real values of x, except x=0.
\(\therefore \) y is a strictly decreasing function for all real values of x(x\(\neq\)0).

Given P= 5(L)^0.7(K)^0.3
Marginal Productivity of Labour (L) is
\({∂P\over ∂L}=5(0.7)(L)^{0.7-1} (K)^{0.3 }= 3.5(L)^{-o·3} (K)^{0.3} = 3.5\left(K\over L\right)^{0.3}\)
When L = 10 and K = 3,
\(\frac { \partial P }{ \partial L } =3.5\left( \frac { 3 }{ 10 } \right) ^{ 0.3 }=3.5(0.3)^{ 0.3 }=3.5\times 0.6968\)
= 2.438 = 2.44
\({∂P\over ∂L}=5(L)^{0.7} (0.3)(K)^{0.3-1 }= 1.5(L)^{0.7}(K)^-{0.7} = 1.5\left(L\over K\right)^{0.7}\)
When L= 10 and K=3,
\({∂P\over ∂L}=1.5\left(10\over3\right)^{0.7}=1.5(3.33)^{0.7} = 1.5(2.2322) = 3.481 = 3.48\)

Let y = x² + 16/x
Differentiating w.r.t. 'x'we get,
\({dy\over dx}=2x-{16\over x^2}\)
Con diittiion for maxi.ma and rmminmimmaa is \({dy\over dx}=0\)
\(2x-{16\over x^2}=0\)
\(⇒2x={16\over x^2}⇒x^2={16\over 2}\)
⇒ x³ = 8 = 2³
⇒ x=2
Also, \({d^2y\over dx^2}=2+{32\over x^3}\)
when x = 2,  \({d^2y\over dx^2}=2+{32\over 8}=2+4=6>0\)
∴ y is minimum when x = 2
∴ Minimum value                 
= 2² + \(16\over 2\)=4+8=12
and there is no maximum.

Let y = x³+8x²+5x-2
Differentiating w.r.t. 'x' we get
\({dy\over dx}=3x^2+16x+5\)
\({dy\over dx}=0⇒3x^2 + 16x + 5=0\)
⇒ (x+5)(3x+1)=0
⇒ x = - 5, -1/3

The possible intervals are (-∞, - 5), (-5, -1/3) and (-1/3, ∞)

Hence the given function is increasing in the intervals (-∞, - 5), (-1/3, ∞) and decreasing in (-5, -1/3).

Let y = x³ + 3x² + 3x + 7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2 + 6x + 3\)
= 3(x² + 2x + 1)
= 3 (x + 1)²
\(\frac { dy }{ dx } >0\) for all real values of x.
\(\therefore\) y is an increasing function for all real values of x.

Given f(x) = x³-3x²-9x+5
Differentiating w.r.t. 'x' we get,
f'(x) = 3x²-6x-9
At stationary points,f'(x) = 0
∴ 3x² - 6x - 9 = 0
⇒ x²-2x-3 = 0 (Divided by 3)
⇒ (x + 1)(x - 3) = 0
The stationary points are obtained when x = -1, x = 3
when x = -1, f(-1) = (-1)³ - 3(-1)² - 9(-1) + 5 = 10
when x = 3, f(3) = (3)³ - 3(3)² - 9(3) + 5 = - 22
∴ The stationary values are 10 and - 22 and the stationary points are (-1, 10) and (3, - 22).