Rules of Differentiation & Comp. Statics Use Cheat Sheet

Constant Function Rule
y = f(x) = k
d/dy(f(x)) = d/dy(k) = 0

Power Rule
y = f(x) = xⁿ
d/dy(f(x)) = d/dy(xⁿ) = n.x^n-1

Power Function Rule Genera­lized
y = f(x) = cxⁿ (where c is the multip­lic­ative constant
d/dy(f(x)) = d/dy(cxⁿ) = cnx^n-1

Sum & Diff Rule
d/dy [f(x) ± g(x)] = d/dy [f(x)] ± d/dy [g(x)] = f '(x) ± g'(x)

Product Rule
d/dy [f(x).g(x)] = f(x).d/dy [g(x)] + g(x).d/dy [f(x)] = f(x).g'(x) + g(x).f '(x)

Finding Marginal Revenue Function from Average Revenue Function
AR = 15 - Q, TR = AR.Q = 15Q-Q², MR = 15 - 2Q
AR = TR/Q = P.Q/Q = P , therefore [Average Revenue is the inverse of the demand as AR is the curve relating Price to output: P = f(Q)]
Under Perfect Compet­ition, the AR curve is horizontal straight line ( MR - AR = ), therefore they must coincide). Under imperfect compet­ition, AR is a downward sloping curve (where MR - AR < 0 for all positive levels of output, MR is above AR)

Quotient Rule
d/dy [f(x)/g(x)] = f '­(x).g(x) + g'(x).f(x)/[g(x)]²

Relati­onship Between Marginal Cost & Average Cost Functions
AC = C(Q)/Q as long as Q>0, The rate of change of AC WRT Q is done by differ­ent­iating AC.
AC' = 1/Q [ C'(Q) - C(Q)/Q], MC = C'(Q)

Exercise 1:
A firm produces ‘q’ units of water. Its total revenue and total cost functions are given as:
TR=-0.5q²+100q and TC=5q²+12q+200

a) Calculate MR and interpret your solution.
MR = dTR/dQ = -q + 100 [It is a decreasing function, therefore a downward sloping curve, MR is the extra revenue that an additional unit of q will bring to the firm, when q = 0, MR is 100 implying that an additional unit of q will decrease the extra revenue]

b) Calculate MC and interpret your solution.
MC = dTC/dQ = 10q + 12 [It is an increasing function, therefore a upward sloping curve, MC is the extra cost that an additional unit of q will bring to the firm, when q = 0, MC = 12 implying that an additional unit of q will increase the extra cost]

c) How many units of water should the firm produce to maximize total profit?
Profit Max = MR = MC
-q+100­=10q+12
11q=88
q=8

Partial Deriva­tives
f 1 = ∂y/∂x1, this is the partial derivative WRT x1, all other indepe­ndent variables in function are held constant

Techniques of Partial Differ­ent­iation
1.Simple Rule
y = f(X1,X2) = 3X²1 + X1X2 + 4X²2
∂y/∂x1 = 6X1+X2; ∂y/∂x2 = X1+8X2
2. Product Rule
y = f(u,v) = (u+4)(­3u+2v)
fu = (u+4)(3) + (3u+2v)(1) = 6u + 2v + 12
fv = (u+4)(2) + (3u+2v)(0) = 2u+8
3.Quotient Rule
y = f(u,v) = (3u-2v)/(u² + 3v)
fu = [(3)(u² + 3v) - (2u)(3­u-2­v)]/[(u² + 3v)]² simplify it
fv = [(-2)(u² + 3v) - (3)(3u­-2v­)]/[(u² + 3v)]² simplify it

A partial derivative is a measure of the instan­taneous rates of change of some variable.
MPPk = Marginal physical Product of Capital, which is the partial derivative Qk relates to the changes of output WRT infini­tesimal changes in capital while labour input is held constant. Similar the partial Derivative of Ql is the mathem­atical repres­ent­ation of the MPPL function.

Exercise 4:
Given the following utility function: f(x,y,z)=x² y³ z 0<x­,y,­z<1
a) Calculate marginal utility with respect to y and interpret.
∂U/∂y = 3x²y²z [MU is + & > 0 for all values of x,y,z, function is strictly monotonic]

Chain Rule
z = f(y), y = g(x); dz/dx = (dz/dy­).(­dy/dx) = f '(y). g'(x)
function can extend to 3 variables:
z = f(y), y = g(x), x = h(w)
dz/dw = (dz/dy­).(­dy/­dx).(d­x/dw) = f '(y). g'(x). h '(w)

Given that TR = f(Q), where Q is a function of labour input, Q = g(L), find dTR/dL, by chain rule, we get dTR/dL = (dTR/d­Q).(­dQ/dL) = f '­(Q).g'(L), [dTR/dQ is the MR function and dQ/dL is the margin­al-­phy­sic­al-­product of Labour (MPPL) function. dTR/dL has the connot­ation of the Marginal revenue product of labour (MRPL)

Exercise 2:
Given TR= PQ and Q=A.K^0.5L^0.3H^0.2

a) Calculate ∂TR/∂K. Interpret your solution.
∂TR/∂K = (dTR/d­Q).(­dQ/dK) = P.[A.0.5.K^-0.5.L^0.3H^0.2]
= 0.5.P.A.L^0.3H^0.2/K^0.5, this is the Marginal Revenue Product of Capital, how does total revenue change when an extra unit of output is produced when one unit of capital is added, dQ/dK is the marginal product of capital (MPK).

b) Calculate ∂Q/∂H. Interpret your solution.
∂Q/∂H = 0.2 A.K^0.5L^0.3/H^0.8, this is the marginal product of Human Capital, it is the extra amount of output that is produced when one unit of Human capital is added, holding all the other inputs constant.

Inverse Function Rule
If a function y=f(x) represents a one-to-one mapping (for every x value, it has a unique y value, vice versa) then f has an inverse f ^-1.

A function is strictly increasing if:
X1>X2 --> f(X1) > f(X2) for all values of x, y

A function is strictly decreasing if:
X1>X2 --> f(X1) < f(X2) for all values of x, y

If a function is either strictly increasing or strictly decreasing it is strictly monotonic.
If a function is monotonic then the function has an inverse.

Thus for example, firms demand curve Q = f(P) that has a negative slope throughout is decrea­sing, as such it has an inverse function, P = f ^-1 (Q) which gives the Av. revenue of the firm P = AR.

For inverse functions, the rule of differ­ent­iation: dx/dy = 1/[dy/dx]
check for strictly monotonic: f’(x) > 0 or f’(x) <0 for all x. If function is a U-shaped curve, may have to check which part of the curve is increasing and which part is decreasing for the x values.

Exercise 3:
Using the TR and TC functions from Exercise 1:

a) Determine whether the total cost function is strictly increa­sing?
TC=5q²+12q+200, if q>0 then TC will increase
MC = 10q+12, it will be positive for all values of q if q>0

b) For what values of q will the total revenue function be strictly increa­sing?
TR=-0.5q²+100q
MR=-q+100 (when q = 0 , MR = 100 which is the turning point for TR) therfore TR will be increasing for q values that are < 100

Market Model
Look at page 170 - 172 in textbook

National - Income Model
Look at page 172 - 173 in textbook, it goes over the income equili­brium and the partial derivative to get the government expend­iture multip­lier, non-income tax multiplier and how an increase in the income tax rate will lower equili­brium income.