Topic 6: Differentiation

Jacques Text Book (edition 4 ): Chapter 4 1.Rules of Differentiation 2.Applications

1 Differentiation is all about measuring change! Measuring change in a linear function: y = a + bx a = intercept b = constant slope i.e. the impact of a unit change in x on the level of y y − y b = ∆y = 2 1 ∆x x2 − x1 2 If the function is non-linear: 40 e.g. if y = x2

20 y=x2

0 0123456X ∆y y − y = 2 1 gives slope of the line ∆x x2 − x1 connecting 2 points (x1, y1) and (x2,y2) on a curve (16-4) • (2,4) to (4,16): slope = /(4-2) = 6

(36-4) 3 • (2,4) to (6,36): slope = /(6-2) = 8 The slope of a curve is equal to the slope of the line (or tangent) that touches the curve at that point

Total Cost Curve

20 y=x2

0 1234567 X which is different for different values of x 4 Example:A firms cost function is Y = X2 X ∆X Y ∆Y 0 0 1 +1 1 +1 2 +1 4 +3

Y = X 3 2 +1 9 +5 4 +1 2 16 +7 Y+∆Y = (X+∆X) 2 2 Y+∆Y =X+2X.∆X+∆X 2 2 ∆Y = +2X. X ∆X+∆X – Y 2 2 since Y = X ⇒ ∆Y = 2X.∆X+∆X ∆ Y ∆ X = 2X+∆X The slope depends on X and ∆X 5 The slope of the graph of a function is called the derivative of the function dy ∆y f '(x) = = lim dx ∆x→0 ∆x • The process of differentiation involves letting the change in x become arbitrarily small, i.e. letting ∆ x → 0 • e.g if = 2X+∆X and ∆X →0 • ⇒ = 2X in the limit as ∆X →0 6 the slope of the non-linear function Y = X2 is 2X • the slope tells us the change in y that results from a very small change in X • We see the slope varies with X e.g. the curve at X = 2 has a slope = 4 and the curve at X = 4 has a slope = 8 • In this example, the slope is steeper

at higher values of X 7 Rules for Differentiation (section 4.3)

1. The Constant Rule If y = c where c is a constant, dy = 0 dx

dy = 0 e.g. y = 10 then dx

8 2. The Linear Function Rule If y = a + bx dy = b dx dy e.g. y = 10 + 6x then = 6 dx

9 3. The Power Function Rule

If y = axn, where a and n are constants dy = n .a .x n −1 dx

dy = 4 x 0 = 4 i) y = 4x => dx

dy 2 = 8 x ii) y = 4x => dx

dy iii) y = 4x-2 => = − 8 x − 3 dx 10 4. The Sum-Difference Rule If y = f(x) ± g(x)

dy d [ f ( x )] d [ g( x )] = ± dx dx dx

If y is the sum/difference of two or more functions of x: differentiate the 2 (or more) terms separately, then add/subtract dy 2 = 4 x + 3 + 3x then dx (i) y = 2x dy = 5 11 (ii) y = 5x + 4 thendx 5. The Product Rule

If y = u.v where u and v are functions of x, (u = f(x) and v = g(x) ) Then

dy dv du = u + v dx dx dx

12 Examples

dy dv du = u + v If y = u.v dx dx dx i) y = (x+2)(ax2+bx) dy = ()x + 2 ()2ax + b + ()ax 2 + bx dx ii) y = (4x3-3x+2)(2x2+4x) dy 3 2 2 =⎜⎛4x −3x+2⎟⎞ ()4x+4 +⎜⎛2x +4x⎟⎞ ⎜⎛12x −3⎟⎞ dx ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 13 6. The Quotient Rule

• If y = u/v where u and v are functions of x (u = f(x) and v = g(x) ) Then du dv v − u dy = dx dx dx v 2

14 du dv v − u u dy If y = then = dx dx v dx v 2

Example 1 ()x + 2 y = ()x + 4 dy ()()()()x + 4 1 − x + 2 1 − 2 = = dx ()x + 4 2 ()x + 4 2

15 7. The Chain Rule (Implicit Function Rule)

• If y is a function of v, and v is a function of x, then y is a function of x and dy dy dv = . dx dv dx

16 dy dy dv = . Examples dx dv dx i) y = (ax2 + bx)½ let v = (ax2 + bx) , so y = v½ 1 dy 1 − = ()ax 2 + bx 2 .()2 ax + b dx 2

ii) y = (4x3 + 3x – 7 )4 let v = (4x3 + 3x – 7 ), so y = v4 dy 3 = 4 ()(4 x 3 + 3 x − 7 . 12 x 2 + 3 ) dx 17 8. The Inverse Function Rule dy 1 = If x = f(y) thendx dx dy • Examples i) x = 3y 2 then dx dy 1 = 6 y = dy so dx 6 y ii) y = 4x3 then dy dx 1 = 12 x 2 = so 2 dx dy 12 x 18 Differentiation in Economics Application I

• Total Costs = TC = FC + VC • Total Revenue = TR = P * Q • π = Profit = TR – TC • Break even: π = 0, or TR = TC • Profit Maximisation: MR = MC

19 Application I: Marginal Functions (Revenue, Costs and Profit) • Calculating Marginal Functions d (TR) MR = dQ d(TC) MC = dQ

20 Example 1

• A firm faces the Solution: demand curve P=17- 3Q TR = P.Q = 17Q – 3Q2 • (i) Find an expression for TR in terms of Q d(TR) MR = =17 − 6Q • (ii) Find an dQ expression for MR in terms of Q

21 Example 2

A firms total cost curve is given by

TC=Q3- 4Q2+12Q (i) Find an expression for AC in terms of Q (ii) Find an expression for MC in terms of Q (iii) When does AC=MC? (iv) When does the slope of AC=0? (v) Plot MC and AC curves and comment on the economic significance of their relationship 22 Solution

(i) TC = Q3 – 4Q2 + 12Q TC 2 Then, AC = / Q = Q – 4Q + 12 d (TC ) = 3Q 2 − 8Q + 12 (ii) MC = dQ

(iii) When does AC = MC? Q2 – 4Q + 12 = 3Q2 – 8Q + 12 ⇒ Q = 2 Thus, AC = MC when Q = 2 23 Solution continued….

(iv) When does the slope of AC = 0? d (AC ) = 2Q − 4 dQ = 0 ⇒ Q = 2 when slope AC = 0 (v) Economic Significance? MC cuts AC curve at minimum point…

24 9. Differentiating Exponential Functions

If y = exp(x) = ex where e = 2.71828…. dy = e x then dx More generally, If y = Aerx dy = rAerx = ry then dx 25 Examples

dy 2x 2x 1) y = e thend x = 2e dy 2) y = e -7x -7x thend x = -7e

26 10. Differentiating Natural Logs

x Recall if y = e then x = loge y = ln y dy • If y = ex then = e x = y dx

• From The Inverse Function Rule dx 1 x = y = e ⇒ dy y • Now, if y = ex this is equivalent to writing x = ln y

dx 1 • Thus, x = ln y ⇒ = dy y 27 More generally, dy 1 if y = ln x ⇒ = dx x NOTE: the derivative of a natural log function does not depend on the co-efficient of x dy 1 Thus, if y = lnmx ⇒ = dx x

28 Proof

• if y = ln mx m>0

• Rules of Logs ⇒ y = ln m+ ln x • Differentiating (Sum-Difference rule) dy 1 1 = 0 + = dx x x

29 Examples dy 1 1) y = ln 5x (x>0) ⇒ = dx x 2) y = ln(x2+2x+1) let v = (x2+2x+1) so y = ln v dy dy dv = . Chain Rule: ⇒ dx dv dx dy 1 = .()2 x + 2 dx x 2 + 2 x + 1 dy (2 x + 2 ) = dx (x 2 + 2 x + 1)

30 3) y = x4lnx Product Rule: ⇒ dy 1 = x 4 + ln x .4 x 3 dx x = x 3 + 4 x 3 ln x = x 3 (1 + 4 ln x ) 4) y = ln(x3(x+2)4) Simplify first using rules of logs ⇒ y = lnx3 + ln(x+2)4 ⇒ y = 3lnx + 4ln(x+2) dy 3 4 = + dx x x + 2

31 Applications II

• how does demand change with a change in price……

•ed= proportional change in demand proportional change in price

∆Q ∆P ∆Q P . = Q P = ∆P Q

32 Point elasticity of demand

dQ P . ed = dP Q ed is negative for a downward sloping demand curve

–Inelastic demand if | ed |<1

–Unit elastic demand if | ed |=1

–Elastic demand if | ed |>1

33 Example 1

-b Find ed of the function Q= aP dQ P e . d = dP Q P e = −baP−b−1. d aP−b −baP−b P = . = −b P aP−b e d at all price levels is –b

34 Example 2 If the (inverse) Demand equation is P = 200 – 40ln(Q+1) Calculate the price elasticity of demand when Q = 20 dQ P . d = dP Q <0 Price elasticity P is of expressed demand: ine terms of Q, dP 40 = − dQ Q + 1 dQ Q + 1 = − Inverse rule ⇒dP 40 Q + 1 P = − . < 0 Hence,d e 40 Q 21 78.22 = − . = -2.05 Q is 20 ⇒d e 40 20 (where P = 200 – 40ln(20+1) = 78.22) 35 Application III: Differentiation of Natural Logs to find Proportional Changes f’(x) The derivative of log(f(x)) ≡ /f(x), or the proportional change in the variable x

i.e. y = f(x), then the proportional ∆ x

dy 1 d (ln y ) . = dx y = dx Take logs and differentiate to find proportional changes in variables 36 dy 1 α α . = 1) Show that if y = x , then dx y x

and this ≡ derivative of ln(y) with respect to x. Solution: dy 1 1 . = .α x α − 1 dx y y

1 x α .α =y x 1 y .α . =y x α

37 Solution Continued…

Now ln y = ln xα Re-writing ⇒ ln y = αlnx d (ln y ) 1 α = α . = ⇒ dx x x Differentiating the ln y with respect to x gives the proportional change in x.

38 Example 2: If Price level at time t is P(t) = a+bt+ct2 Calculate the rate of inflation.

Solution: Alternatively, The inflation rate at t is the proportional change in p differentiating the log of P(t) wrt t directly lnP(t) = ln(a+bt+ct 2 )

1 dP(t) b+2ct where v = (a+bt+ct2 . = ) so lnP = ln v P(t) dt a bt ct2 + + Using chain rule,

d(lnP(t )) b+2ct = dt a +bt +ct2