Lecture for Week 4 (Secs. 3.1–3)Derivatives(Finally, we get to the point.)1There are two equivalent forms of the defini-tion of a derivati ve:f′(a) = limx→af(x) − f(a)x − a.f′(x) = limh→0f(x + h) − f(x)h.One uses whichever seems most natural or usefulin a given context. Usually the second is moreconvenient for calculating derivatives from firstprincip les.2I assume that you have read all the explana-tions in the book, so I will most ly do examples.Exercise 3.1.27Find the derivati ve of f(x) = x4using the defini-tion of derivative.Actually, we already did this, for the specialcase x = 1, remember?3f′(x) = limh→0(x + h)4− x4h.I showed you how to do this in Lecture 2, slides25–27.(x + h)4= x4+ 4x3h + O(h2)where O(h2) is shorthand for “terms contai ning4h2or even higher powers”. Sof′(x) = limh→04x3h + O(h2)h= limh→04x3+ O(h)= 4x3.5Now I hear some of you saying, “Why arewe doing this? Don’t we all know that the deriv-ative of xnis nxn−1? And that easy formula isin t he next sec t ion.”The i mmed i at ely p ract ical reason is t hatyou ’ll have to do something like this on the firstexam, an d maybe also on the final, so you’d bet-ter get used to it.The deeper reason is that formulas are use-6less unless you know what they mean . In sci-ence and engi neering courses you’ll study howequations invol ving derivative s arise out of thephysics of a problem, by considerin g how somevariabl es change as other variables ch ange. Therethe derivative is the answer, not the question. Inthose situations you don ’t know beforeh and whatthe functions i nvolved are; you j ust have to un-derstand t hat one is the derivative of anothe r.Applying the definition t o particular fun ctionsis si mpl e, less abstract, practice for that kind of7thinking.Also, of course, we will need the d efinition,and t he skills of workin g wit h it, to fin d deriva-tives of non polynomial functions such as sin xlater.So, l et’s d o one more, pretending that n oneof you have eve r heard of the chain rule and thepower rule.8Exercise 3.1.23Find the derivative of g(x) =√1 + 2x using thedefini t ion of derivative.9g′(x) = limh→0√1 + 2x + 2h −√1 + 2xh.Somehow we have to squeeze an h out of the nu-merator so we can cance l it. Experience showsthat multiplying top an d bottom by√1 + 2x + 2h +√1 + 2x10might help. (See Lecture 3, sl i de 14.)g′(x) = limh→0(1 + 2x + 2h) − (1 + 2x)h(√1 + 2x + 2h +√1 + 2x)= limh→02√1 + 2x + 2h +√1 + 2x=1√1 + 2x.11Ok, l et’s go on to the next section and usethe powe r rule and other theorems!Exercise 3.2.7Differentiate Y (t) = 6t−9.12This is a simple in s t ance of th e power rule:Y′(t) = (−9)6t−10= −54t−10(alias −54t10).13Exercise 3.2.15Differentiatey =x2+ 4x + 3√x.14Hard way: Use the qu ot i ent rule.Easy way:√x = x1/2, soy = x3/2+ 4x1/2+ 3x−1/2.y′=32x1/2+ 2x−1/2−32x−3/2.If you like, you can write this back in terms ofsquare roots, as in the book’s answer.15Exercise 3.2.37Find the equ ation of the t angent line to thecurvey = x +4xat the point (2, 4).16The slope of the line is the derivative of thefunct i on.y′= 1 −4x2.y′(2) = 1 −11= 0.So th e tangent line i s the horizontal lin e throughthe point:y = 4.17Usually the tangent will not be horizontal,so you have to use the point-slope form of a line:y = f′(a)(x − a) + f ( a)is the tangent line to the graph of f at the point(a, f(a)).18If g is t he derivative of f, t hen f is called anantiderivative of g.Later we will see that any two antideriva-tives of g differ on ly by a con s t ant (providedthat t he domain of the functions is the wholereal li ne, or a single interval).You need to find antiderivatives in first-semester physics, so we will treat them brieflynow.19Exercise 5.7.1 (p. 353)Find the most ge neral antiderivative off(x) = 12x2+ 6x − 5.Exercise 5.7.9Find the most ge neral antiderivative ofg(t) =t3+ 2t2√t.20f(x) = 12x2+ 6x − 5.Use the power rule in reverse , remembering todivide each term by the new exponent:F (x) = 4x3+ 3x2− 5x + C(C = arbitrary constant) is the most generalantiderivative of f .As you may know already, later we’ll writeZ(12x2+ 6x − 5) dx = 4x3+ 3x2− 5x + C.21g(t) =t3+ 2t2√t= t5/2+ 2t3/2.SoG(t) =27t7/2+45t5/2+ Cis the most gen eral antiderivative.Replacing the square root factor by a power,which was convenient in an earlier example (slide15), is almost essential now.22Exercise 3.3.7The position function of a particle iss = t3− 4.5t2− 7t (for t ≥ 0).When d oes the particle reach a velocity of 5m/s?23s = t3− 4.5t2− 7t (for t ≥ 0).The velocity is t he derivative of the position:v = 3t2− 9t − 7.5 = 3t2− 9t − 7 ⇒ 3t2− 9t − 12 = 0.0 = t2− 3t − 4 = (t − 4)(t + 1).The root t = −1 is outsi de the s t ated domain , sot = 4 s.24Exercise 3.3.11A s t one is dropped into a lake, creating a circu -lar ripple that travels outward at a speed of 60cm/s. Find the rate at wh i ch the circle’s area isincreasing after(a) 1 s(b) 3 s(c) 5 s25We need the formul a for th e area of a cir-cle of radius r: A = πr2. We also need to knowwhat the radius is at each time ; from the prob-lem statement, th at is clearly r = 60t. Therefore,A = (60)2πt2= 3600πt2.dAdt= 7200πt.To get the numerical answers, ins ert t = 1, 3, 5.(Units = cm2/s
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