Lesson 23: The Chain Rule

Slide 1: Lesson 23 (Sections 16.1–2) The Chain Rule Math 20 November 14, 2007 Announcements Problem Set 9 assigned today. Due November 21. There will be class November 21. next OH: Today 1-3pm Midterm II: 12/6, 7-8:30pm in Hall A.

Slide 2: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 3: HW problem 15.7.2 Recall that a discriminating monopolist can choose the price/quantity sold in two markets. P1 = a1 − b1 Q1 P2 = a2 − b2 Q2 Suppose cost increases constantly with quantity: C = α(Q1 + Q2 ). The profit is therefore π = P1 Q1 + P2 Q2 − α(Q1 + Q2 ) = (a1 − b1 Q1 )Q1 + (a2 − b2 Q2 )Q2 − α(Q1 + Q2 ) 2 2 = (a1 − α)Q1 − b1 Q1 + (a2 − α)Q2 − b2 Q2

Slide 4: Solution Completing the square gives 2 2 π = (a1 − α)Q1 − b1 Q1 + (a2 − α)Q2 − b2 Q2 2 (a1 − α) (a1 − α)2 = −b1 Q1 − + 2b1 4b1 2 (a2 − α) (a2 − α)2 − b2 Q 2 − + 2b2 4b2 The optimal quantities are ∗ a1 − α ∗ a2 − α Q1 = Q2 = 2b1 2b2

Slide 5: The corresponding prices are ∗ a1 + α ∗ a2 + α P1,d = P2,d = 2 2 The maximum profit is ∗ (a1 − α)2 (a2 − α)2 πd = + 4b1 4b2

Slide 6: The Indiscriminating Monopolist An indiscriminating monopolist can only set one price each each “area”: So P = a1 − b1 = a2 − b2 Q2 . This means we can get Q1 and Q2 , and therefore π, all in terms of P: a1 − P a2 − P Q1 = Q2 = b1 b2 π(P) = P(Q1 + Q2 ) − α(Q1 + Q2 ) a1 − P a2 − P a1 − P a2 − P =P + −α + b1 b2 b1 b2 a1 + α a2 + α 1 1 αa1 αa2 = + P− + P2 − − b1 b2 b1 b2 b1 b2

Slide 7: Complete the square in P now and we get a1 +α a2 +α b1 + b2 b2 b1 Pi∗ = = ∗ P1,d + P∗ 2 1 + 1 b1 + b2 b1 + b2 2,d b1 b2 2 a1 +α a2 +α b1 + b2 αa1 αa2 πi∗ = − − 4 1 + 1 b1 b2 b1 b2

Slide 8: Complete the square in P now and we get a1 +α a2 +α b1 + b2 b2 b1 Pi∗ = = ∗ P1,d + P∗ 2 1 + 1 b1 + b2 b1 + b2 2,d b1 b2 2 a1 +α a2 +α b1 + b2 αa1 αa2 πi∗ = − − 4 1 + 1 b1 b2 b1 b2 Subtract and do some algebra: (a1 − a2 ) 2 πd − πi∗ = ∗ >0 4 (b1 + b2 )

Slide 9: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 10: Optimization of functions of several variables Last week: algebraic optimization Critical values of quadratic forms This week: more differentiation Chain Rule Implicit differentiation Next week: unconstrained optimization Approximate functions “to second order” and use rules for quadratic forms Following week: constrained optimization

Slide 11: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 12: The Chain Rule in one variable See Section 5.2 for more (f ◦ g ) (x) = f (g (x)) · g (x)

Slide 13: The Chain Rule in one variable See Section 5.2 for more (f ◦ g ) (x) = f (g (x)) · g (x) Or, if y = f (u) and u = f (x), then dy dy du = dx du dx

Slide 14: The Chain Rule in one variable See Section 5.2 for more (f ◦ g ) (x) = f (g (x)) · g (x) Or, if y = f (u) and u = f (x), then dy dy du = dx du dx A goal for today is the chain rule in several variables.

Slide 15: Example Let z = xy 2 , and suppose x and y are given as functions of t: x = t3 y = sin t dz Find dt .

Slide 16: Example Let z = xy 2 , and suppose x and y are given as functions of t: x = t3 y = sin t dz Find dt . Solution z = t 3 sin2 t, so dz = 3t 2 sin2 t + t 3 2 sin t cos t dt dx y2 x dy dt dt

Slide 17: Fact (The Chain Rule, version I) When z = F (x, y ) with x = f (t) and y = g (t), then z (t) = F1 (f (t), g (t))f (t) + F2 (f (t), g (t))g (t) or dz ∂F dx ∂F dy = + dt ∂x dt ∂y dt

Slide 18: Fact (The Chain Rule, version I) When z = F (x, y ) with x = f (t) and y = g (t), then z (t) = F1 (f (t), g (t))f (t) + F2 (f (t), g (t))g (t) or dz ∂F dx ∂F dy = + dt ∂x dt ∂y dt We can generalize to more variables, too. If F is a function of x1 , x2 , . . . , xn , and each xi is a function of t, then dz ∂F dx1 ∂F dx2 ∂F dxn = + + ··· + dt ∂x1 dt ∂x2 dt ∂xn dt

Slide 19: Tree Diagrams for the Chain Rule F ∂F ∂F ∂x ∂y x y dx dy dt dt t t To differentiate with respect to t, find all “leaves” marked t. Going down each branch, chain (multiply) all the derivatives together. Then add up the result from each branch. dz dF ∂F dx ∂F dy = = + dt dt ∂x dt ∂y dt

Slide 20: Example Consider a Cobb-Douglas production function defined by P(A, x, y ) = AK a Lb where K is capital, L is labor, and A is the “technology” to convert these quantities to production. Suppose that all of these are changing over time. Show that 1 dP 1 dA 1 dK 1 dL = +a +b P dt A dt K dt L dt That is relative rate relative rate relative rate relative rate of growth in = of growth of + a × of growth of + b × of growth of output technology capital labor

Slide 21: Solution dP ∂P dA ∂P dK ∂P dL = + + dt ∂A dt ∂K dt ∂L dt dA dK dL = K a Lb + AaK a−1 Lb + AK a bLb−1 dt dt dt So 1 dP K a Lb dA + AaK a−1 Lb dK + AK a bLb−1 dL dt dt dt = P dt AK a Lb 1 dA 1 dK 1 dL = +a +b A dt K dt L dt

Slide 22: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 23: Fact (The Chain Rule, Version II) When z = F (x, y ) with x = f (t, s) and y = g (t, s), then ∂z ∂F ∂x ∂F ∂y = + ∂t ∂x ∂t ∂y ∂t ∂z ∂F ∂x ∂F ∂y = + ∂s ∂x ∂s ∂y ∂s F x y t s t s

Slide 24: Example ∂z ∂z Suppose z = xy 2 , x = t + s and y = t − s. Find ∂t and ∂s at (t, z) = (1/2, 1) in two ways: (i) By expressing z directly in terms of t and s before differentiating. (ii) By using the chain rule.

Slide 25: Theorem (The Chain Rule, General Version) Suppose that u is a differentiable function of the n variables x1 , x2 , . . . , xn , and each xi is a differentiable function of the m variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and ∂u ∂u ∂x1 ∂u ∂x2 ∂u ∂xn = + + ··· + ∂ti ∂x1 ∂ti ∂x2 ∂ti ∂xn ∂ti

Slide 26: Theorem (The Chain Rule, General Version) Suppose that u is a differentiable function of the n variables x1 , x2 , . . . , xn , and each xi is a differentiable function of the m variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and ∂u ∂u ∂x1 ∂u ∂x2 ∂u ∂xn = + + ··· + ∂ti ∂x1 ∂ti ∂x2 ∂ti ∂xn ∂ti In summation notation n ∂u ∂u ∂xj = ∂ti ∂xj ∂ti j=1

Slide 27: Example Write out the Chain Rule for the case where w = f (x, y , z, t) and x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ).

Slide 28: Example Write out the Chain Rule for the case where w = f (x, y , z, t) and x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ). w x y z t u v u v u v u v

Slide 29: Example Write out the Chain Rule for the case where w = f (x, y , z, t) and x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ). w x y z t u v u v u v u v Solution ∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t = + + + ∂u ∂x ∂u ∂y ∂u ∂z ∂u ∂t ∂u ∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t = + + + ∂v ∂x ∂v ∂y ∂v ∂z ∂v ∂t ∂v

Slide 30: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 31: Matrix Perspective ∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t = + + + ∂u ∂x ∂u ∂y ∂u ∂z ∂u ∂t ∂u ∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t = + + + ∂v ∂x ∂v ∂y ∂v ∂z ∂v ∂t ∂v Or,  ∂x ∂x   ∂u ∂v   ∂y ∂y  ∂w ∂w ∂w ∂w ∂w ∂w   ∂u  ∂v  =  ∂z ∂z  ∂u ∂v ∂x ∂y ∂z ∂t    ∂u ∂v    ∂t ∂t ∂u ∂v

Slide 32: Outline HW problem 15.7.2 Where we’re going Chain Rule I Chain Rule II Matrix expressions for the Chain Rule Leibniz’s Formula for Integrals

Slide 33: Leibniz’s Formula for Integrals Fact Suppose that f (, t, x), a(t), and b(t) are differentiable functions, and let b(t) F (t) = f (t, x) dx a(t)

Slide 34: Leibniz’s Formula for Integrals Fact Suppose that f (, t, x), a(t), and b(t) are differentiable functions, and let b(t) F (t) = f (t, x) dx a(t) Then b(t) ∂f (t, x) F (t) = f (t, b(t))b (t) − f (t, a(t))a (t) + dx a(t) ∂t

Slide 35: Leibniz’s Formula for Integrals Fact Suppose that f (, t, x), a(t), and b(t) are differentiable functions, and let b(t) F (t) = f (t, x) dx a(t) Then b(t) ∂f (t, x) F (t) = f (t, b(t))b (t) − f (t, a(t))a (t) + dx a(t) ∂t Proof. Apply the chain rule to the function v H(t, u, v ) = f (t, x) dx u with u = a(t) and v = b(t).

Slide 36: Tree Diagram H t u v t t

Slide 37: More about the proof v H(t, u, v ) = f (t, x) dx u

Slide 38: More about the proof v H(t, u, v ) = f (t, x) dx u Then by the Fundamental Theorem of Calculus (see Section 10.1) ∂H ∂H = f (t, v ) = −f (t, u) ∂v ∂u

Slide 39: More about the proof v H(t, u, v ) = f (t, x) dx u Then by the Fundamental Theorem of Calculus (see Section 10.1) ∂H ∂H = f (t, v ) = −f (t, u) ∂v ∂u Also, v ∂H ∂f = (t, x) dx ∂t u ∂x since t and x are independent variables.

Slide 40: Since F (t) = H(t, a(t), b(t)), dF ∂H ∂H du ∂H dv = + + dt ∂t ∂u dt ∂v dt b(t) ∂f = (t, x) + f (t, b(t))b (t) − f (t, a(t))a (t) a(t) ∂x

Slide 41: Application Example (Example 16.8 with better notation) Let the profit of a firm be π(t). The present value of the future profit π(τ ) where τ > t is π(τ )e −r (τ −t) , where r is the discount rate. On a time interval [0, T ], the present value of all future profit is T V (t) = π(τ )e −r (τ −t) dt. t Find V (t).

Slide 42: Application Example (Example 16.8 with better notation) Let the profit of a firm be π(t). The present value of the future profit π(τ ) where τ > t is π(τ )e −r (τ −t) , where r is the discount rate. On a time interval [0, T ], the present value of all future profit is T V (t) = π(τ )e −r (τ −t) dt. t Find V (t). Answer. V (t) = rV (t) − π(t)

Slide 43: Solution Since the upper limit is a constant, the only boundary term comes from the lower limit: T ∂ V (t) = −π(t)e −r (t−t) + π(τ )e −r τ e rt dτ t ∂t T = −π(t) + r π(τ )e −r τ e rt dτ t = rV (t) − π(t). This means that π(t) + V (t) r= V (t) So if the fraction on the right is less than the rate of return for another, “safer” investment like bonds, it would be worth more to sell the business and buy the bonds.