MATH 234, Calculus for BusinessLecture 8, Textbook Sections 4.1More Examples of DerivativesUniversity of Illinois, Urbana-ChampaignFebruary 13th, 2017More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 1 / 20AnnouncementsThere are many different notations for derivatives; be careful.Today is the last day to be added to the list for the conflict exam.Office hours are changed: Monday, 2/13, 11am to 12pm and also3pm to 4pm in 24 Illini Hall. No office hours on Wednesday, 2/15.Tutoring hours are changed for this week only (all in 441 Altgeld Hall):Day Time Teaching AssistantsMonday 4:00 to 6:00pm William Karr, Dara Zirlin6:00pm to 7:00pm Christopher Gartland7:00pm to 9:00pm Mingyu ZhaoTuesday 4:00 to 6:00pm Dileep Menon, Paulina Koutsaki6:00 to 8:00pm Artur Kirkoryan, Alessandro Gondoloand Lutian ZhaoWed. and Thurs. tutoring hours are cancelled.Thurs. section is cancelled, but Tues. section is still ON.More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 2 / 20ReviewDefinitionThe instantaneous rate of change of f(x) at x = a isf0(a) = limh→0f (a + h) − f (a)(a + h ) − a= limh→0f (a + h) − f (a)hThis is also called the derivative of f(x) at x = a.DefinitionThe tangent line to the graph of y = f (x) at the point (a, f (a ))is the line through (a, f (a)) with slope f0(a).The equation of this tangent line is y − f (a) = f0(a) ∗ (x − a)Definition (The operation of differentiation, Version 1)The derivative of f (x) is the functionf0(x) = limh→0f (x + h) − f (x)hNote: The derivative inputs a function and outputs another function.More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 3 / 20Section 4.1, Techniques for Finding Derivatives Three Definitions (all are the same)Definition (The operation of differentiation, Version 1: Lagrange, 1793)The derivative of f (x) is the functionf0(x) = limh→0f (x + h) − f (x)hDefinition (The operation of differentiation, Version 2: Leibniz, 1675)The derivative of f (x) is the functiondfdx(x) = limh→0f (x + h) − f (x)horddx[f (x)] = limh→0f (x + h) − f (x)hDefinition (The operation of differentiation, Version 3: Newton, 1666)The derivative of f (x) is the function˙f (x) = limh→0f (x + h) − f (x)hNote: The derivative inputs a function and outputs another function.Even more (10+) notations exist; people argue notations to this day.More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 4 / 20Section 4.1, Techniques for Finding Derivatives ExamplesExampleLet f (x) = 2 for all x. Computedfdx(x).Solutiondfdx(x) = limh→0f (x + h) − f (x)h= limh→02 − 2h= limh→00 = 0This means the instantaneous rate of change is always 0.-4 -2 2 41234More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 5 / 20Section 4.1, Techniques for Finding Derivatives ExamplesTheorem (Constant function = no change of output)Let k be a constant, real number and let f (x) = k for all x.Then f0(x) = 0 for all x.Similar to before, the proof isf0(x) = limh→0f (x + h) − f (x)h= limh→0k − kh= limh→00 = 0More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 6 / 20Section 4.1, Techniques for Finding Derivatives ExamplesExampleLet f (x) = 3x + 5. Compute f0(x) using the definition of the derivative.Solutionf0(x) = limh→0f (x + h) − f (x)h= limh→03 ∗ (x + h) + 5− (3 ∗ x + 5)h= limh→03x + 3h + 5− (3x + 5)h= limh→03hh= limh→03 = 3More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 7 / 20Section 4.1, Techniques for Finding Derivatives ExamplesTheorem (Instantaneous rate of change on a line is the slope of the line)Let f (x) = mx + b (m, b are constant). Then f0(x) = m.Similar to before, the proof isf0(x) = limh→0f (x + h) − f (x)h= limh→0m ∗ (x + h) + b− (m ∗ x + b )h= limh→0mx + mh + b− (mx + b)h= limh→0mhh= limh→0m = mMore Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 8 / 20Section 4.1, Techniques for Finding Derivatives ExamplesExampleLet f0(1) = 2 and g0(1) = 3. Define T (x) = 4f (x) + 5g(x).Compute T0(1) using the limit laws and definition of the derivative.SolutionT0(1) = limh→0T (1 + h) − T (1)h= limh→04f (1 + h) + 5g(1 + h)−4f (1) + 5g(1)h= limh→04f (1 + h) − 4f (1) + 5g(1 + h) − 5g(1)h= limh→04 ∗f (1 + h) − f (1)h+ 5 ∗g(1 + h) − g(1)h= 4 ∗ limh→0f (1 + h) − f (1)h+ 5 ∗ limh→0g(1 + h) − g(1)h= 4 ∗ f0(1) + 5 ∗ g0(1) = 4 ∗ 2 + 5 ∗ 3 = 23More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 9 / 20Section 4.1, Techniques for Finding Derivatives ExamplesTheoremLet f (x) and g(x) be differentiable functions (i.e., their derivatives exists).If T (x) = af (x) + bg(x), then T0(x) = af0(x) + bg0(x) (a, b are constant).Similar to before, the proof isT0(x) = limh→0T (x + h) − T (x)h= limh→0af (x + h) + bg(x + h)−af (x) + bg(x)h= limh→0af (x + h) − af (x) + bg(x + h) − bg(x)h= limh→0a ∗f (x + h) − f (x)h+ b ∗g(x + h) − g(x)h= a ∗ limh→0f (x + h) − f (x)h+ b ∗ limh→0g(x + h) − g(x)h= a ∗ f0(x) + b ∗ g0(x)More Examples of Derivatives (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business February 13th, 2017 10 / 20Section 4.1, Techniques for Finding Derivatives Properties of DerivativesTheoremLet f (x) and g(x) be differentiable functions (i.e., their derivatives exists).If T (x) = af (x) + bg(x), then T0(x) = af0(x) + bg0(x) (a, b are constant).Theorem (Using a = 1 and b = 1: The Sum Rule)Let f (x) and g(x) be differentiable functions (i.e., their derivatives exists).If T (x) = f (x) + g(x), then T0(x) = f0(x) + g0(x)Theorem (Using a = 1 and b = −1: The Difference Rule)Let f (x) and g(x) be differentiable functions (i.e., their derivatives exists).If T (x) = f (x) − g (x), then T0(x) = f0(x) − g0(x)Theorem (Using a = k and b = 0: The Constant Multiplication Rule)Let f (x) be a differentiable function and let k be constant.If T (x) = kf