SOME DIFFERENTIAL CALCULUS YOU MAY NEED 1. THE FIRST DERIVATIVE 1.1 For a power function y = f(x) = Axm , dy/dx = Amxm-­‐1 This holds for ALL values of m. If m= 0, i.e.-­‐ y = Ax0 (which is equal to A) dy/dx = A0x0-­‐1 = 0. So the derivative of a fixed (constant) quantity is zero Examples: a. For a linear function of a single variable: y = f(x) = ax dy/dx = a Thus for a cost function C(q) = 4q, the marginal cost dC/dq = 4 b. For an affine (linear with intercept) function: y = f(x) = c + mx dy/dx = m (note that the c vanishes) For a cost function with a fixed cost, C(q) = 180 + 3q, dC/dq = 3 c. We can also get a function of mixed higher order polynomials y = f(x) = a + bx + cx2+ dx3 dy/dx = b + 2cx + 3dx2 For a cost function C(q) = 150 + 34q3 + 20q2, dC/dq = 102q2 + 40q For an inverse demand function, p = 10 – 4q, the total revenue TR = pq TR = q (10-­‐4q) = 10q – 4q2 Thus Marginal Revenue = MR = d(TR)/dq = 10 – 8q 1.2 For a logarithmic function to base e y = f(x) = aln(x), dy/dx = a/x y = 2ln(x), dy/dx = 2/x Example: If quantity demanded is semi-­‐log, i.e.-­‐ q = 10-­‐ 2ln(p), then dq/dp = -­‐2/p

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If quantity demanded is log-­‐log, i.e.-­‐ ln(q) = 10 -­‐ 2 ln(p) dln(q) = -­‐ 2dln(p) Thus dln(q)/dln(p) = -­‐2. Notice that the LHS is simply the price elasticity of demand, which is 2 (or -­‐2 depending on whether you choose to keep or omit the – sign). 1.3 For a product function y = u(x).v(x) dy/dx = u(x).[dv(x)/dx] + v(x).[du(x)/dx] Example: TR = p(q).q MR = dTR/dq = p(q).[dq/dq] + q.[dp(q)/dq] = p(q).[1] + q[dp(q)/dq]

= p(q) + qdp(q)/dq 1.4 For a quotient function y = u(x)/v(x) = {v(x).[du(x)/dx] – u(x).[dv(x)/dx]}/[v(x)]2 Example: AC(q) = C(q)/q dAC(q)/dq = {q[dC(q)/dq]– C(q).[dq/dq]}/q2 = (1/q)*{[dC(q)/dq] – (C(q)/q)} = (1/q)*[MC – AC] 2. THE SECOND DERIVATIVE Is obtained by taking a derivative of the first derivative For y = f(x) = Axm , dy/dx = Amxm-­‐1 , d2y/dx2 = Am(m-­‐1)xm-­‐2 Example: Suppose production in the SR is according to the production function: y= f(K*, L) = 2K*1/2L1/2 (Note that capital is fixed in the SR) df/dL = K*1/2L-­‐1/2 (this is MPL) and d2f/dL2 = (-­‐1/2)*K*1/2L-­‐3/2 (notice that MPL is diminishing) 3. DERIVATIVE OF A FUNCTION OF MORE THAN ONE VARIABLE Suppose y = f(a, b)

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Then taking a derivative with respect to one variable, for example a, would mean that we treat the other variable (b) as a constant. We would treat a as constant when we take the derivative with respect to b The derivatives ∂y/∂a and ∂y/∂b are known as “partial” derivatives. Example: Utility function U(x, y) = 2x + 3y (perfect substitutes) ∂U/∂x = 2, ∂U/∂y = 3. These are the marginal utilities. Utility function U(x, y) = 2x1/2y1/2 Now the marginal utilities are, ∂U/∂x = x-­‐1/2y1/2, ∂U/∂y = x1/2y-­‐1/2 Notice that we treated y as a constant in the first expression and x as constant in the second expression. We can take a second derivative by using the same rules. For U(x, y) = 2x + 3y, ∂2U/∂x2 = 0, ∂2U/∂y2 = 0 For U(x, y) = 2x1/2y1/2 , ∂2U/∂x = (-­‐1/2)*x-­‐3/2y1/2, ∂2U/∂y = (-­‐1/2)*x1/2y-­‐3/2 3.1 Total differentiation Finally, when both a and b change for the function y = f(a, b), the expression for the change in the value of the function is given by: dy = [∂y/∂a]*da + [∂y/∂b]*db The partial of y with respect to a times the change in a, plus the partial of the change in y with respect to b times the change in b. If we had more variables we would just extend the right hand side. Example: For U(x, y) = 2x + 3y, dU = 2*dx + 3*dy. If you recall, that along an indifference curve dU = 0, thus the MRS = -­‐dy/dx = 2/3 4. Small and large changes Finally, when we consider infinitesimal (very small) changes, we use ∂ or d. This depends on whether the function has one argument (use d) or whether it has more than one argument (use ∂). When we speak of finite (larger) changes, we use ∆. The derivative represents the slope of a line or a curve. When we use larger changes, we define slope of a line segment. When we use small changes we define slope at a point.

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