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).