Exercise 1 – Find the first derivative and the second derivative of the following functions
Answer: Applying constant function and power function rule
(A) Y = 3 + 10X + 5X2 dY/dX = 0 + 1.10.X1-1 +2.5.X2-1 dY/dX = 10 + 10X d2Y/ dX2 = 0 + 1.10.X1-1 d2Y/ dX2 = 10
(B) Y = 2X (4 + X3 )
Y = 8X + 2X4 dY/dX = 1.8.X1-1 + 4.2.X4-1 dY/dX = 8 + 8X3 d2Y/ dX2 = 0 + 3.8.X3-1 d2Y/ dX2 = 24X2
(C) Y = 3 /X2
Y = 3X-2 dY/dX = -2.3.X-2-1 dY/dX = -6X-3 dY/dX = -6/X3 d2Y/ dX2 = -3.-6X-3-1 d2Y/ dX2 = 18X-4 d2Y/ dX2 = 18/X4

(D) Y = 18T – 2T2 dY/dT = 1.18.T1-1 – 2.2.T2-1 dY/dT = 18 – 4T d2Y/ dT2 = 0 – 1.4.T1-1 d2Y/ dT2 = - 4 Exercise 2 - Find the partial Derivative of Y with respect to X Answer
(A) Y = 10 + 3Z + 2X
∂Y/∂X = 0 + 0 + 1.2.X1-1
∂Y/∂X = 2
(B) Y= 18Z + X2 + Z.X
∂Y/∂X = 0 + 2.1.X2-1 + Z
∂Y/∂X = 2X + Z

Application - The nursing home industry is growing rapidly because the aging of American population. According to the study of an economist, the average cost per patient day of a nursing home can be approximated by
C = A – 0.16B + 0.002B2
Where, B is the nursing home’s number of patient days per year ( in thousands) and A is the number that depends on the location and other factors but not on B. Based on the information , how big must a nursing home be ( in terms of patient – days) to minimize the cost per patient day ?
Answer – C= ƒ (A, B)
Where C is Avg. Cost per patient day
A is variable depends on location and other factors
B is number of patient days per year
∂C/∂A = 1.A1-1 + 0 + 0
∂C/∂A = 1
∂2C/∂A2 = 0
∂C/∂B = 0 – 0.16 + 0.004B
∂C/∂B = – 0.16 + 0.004B
Equating ∂C/∂B = 0
– 0.16 + 0.004B = 0
B = 40
Second derivative
∂2C/∂B2 = 0 + 0.004
∂2C/∂B2 = 0.004 > 0
Since the second derivative ∂2C/∂B2 is positive and ∂2C/∂A2 is constant therefore C reaches a minimum.
Conclusion: A nursing home should achieve