Math 221 – Exam I (50 minutes) – Friday Feb 22, 2002AnswersI. (30 points.) Find the indicated derivative.(a) y =µx + 1x − 1¶2.dydx=?Answer: By the chain rule and the quotient ruledydx= 2µx + 1x − 1¶·1(x − 1) − (x + 1)1(x − 1)2.(This is problem 8 on page 75 of the text.)(b) x =t1 + tand y =t21 + t.dydx=?Answer: By the quotient ruledxdt=(1 + t) − t(1 + t)2=1(1 + t)2,dydt=2t(1 + t) − t2(1 + t)2=t2+ 2t(1 + t)2so by the chain ruledydx=dy/dtdx/dt= t2+ 2t.From x = 1/(1 + t) we get x + tx = t and then t = x/(1 − x) sodydx=µx1 − x¶2+ 2µx1 − x¶.(This is problem 4 on page 93 of the text.)(c) y = sec25x = (cos 5x)−2.dydx=?Answer: By the chain ruledydx= −2(cos 5x)−3ddxcos(5x) = −2(cos 5x)−3(−5 sin 5x).(This is Example 4 on page 105 of the text.)Grader’s commentsSome students did not have good understanding of the chain rule andhence wrote completely wrong answers for (a) and (c). When the chain rule1was followed by a non-trivial differentiation as in (a) and (c), many studentsfailed to use appropriate differentiation rules.In (b), many students wrote too much which made their work hard tofollow. I was also able to see the fact that they don’t have to simplify theiranswers has some undesirable effects:1. Many people solved for t in x and substituted that to express y in termsof x and then tried to find dy/dx. But while they were doing that, theydid not simplify anything in the process hence applying the quotientrule many times inside the quotient rule. It was very hard to followand most of those students ended up making mistakes.2. In some cases, there were obvious cancellations and they just simplywouldn’t cancel them. Again that made it very hard to follow theiranswers. Things like t(−1 · (1 + t−1)−2) or −2(cos5x)−3(−sin5x · 5) asfinal answers were not visually pleasing to see.I think it’s a good policy that they don’t have to simplify expressions whenthere’s absolutely no benefit of doing that. But I think they should simplifyto reasonable extent. I also personally think that they should be capable ofsimple algebra.Lastly, some students wrote −sin in their answer hence exhibited thatthey do not have good understanding of cosine or sine functions.Professor’s responseI have wrestled with the problem of how much simplification and what al-gebraic skills to require ever since I began teaching calculus over thirty yearsago. My problem is that I don’t know how to word a question which forcessimplification. My current approach is to tell students that they needn’t sim-plify unless this is necessary to complete the problem, but I certainly do believethat algebra skills are important. Any ideas?II. (20 points.) Find an equation for the tangent line to the curvex2+ xy − y2= 1at the point P0(2, 3).Answer: By implicit differentiation2x + y + xdydx− 2ydydx= 02sodydx=2x + y2y − x.The slope of the tangent line at P0isdydx¯¯¯¯(x,y)=(2,3)=4 + 36 − 2=74and the equation of the tangent line isy − 3 =74(x − 2)(This is Problem 30 on page 83 of the text.)Grader’s CommentsStudents did very well on this problem, almost all of them got 20 or alittle less in case they had an arithmetical or notational mistake. I assigned10/20 for finding the derivative and 5/10 for each of finding the slope andthe equation. The following error was common: dy/dx = 2x + y + x dy/dx +2y dy/dx = 0, I took off a point for that.III. (30 points.) Two functions f and g are inverse functions if the equationsy = f(x) and x = g(y) have the same graph, i.e. if x = g(f(x)) andy = f(g(y)). What is the inverse function g of the function f(x) = x3+ 1and what is the derivative of g?Answer: Since y = x3+ 1 ⇐⇒ x = (y − 1)1/3The inverse function isg(y) = (y − 1)1/3sog0(y) =(y − 1)−2/33.This is a consequence of the Chain Rule: g(f(x)) = x so g0(f(x))f0(x) = 1so if y = f(x), theng0(y) = g0(f(x)) =1f0(x)=13x2=13(y − 1)2/3.(This is Example 3 on page 78.)Remark. The function g defined by the formulag(y) = (y − 1)133is the same as the function g defined by the formulag(x) = (x − 1)13.Grader’s CommentsMany students (probably because they were taught to in an earlier course)immediately went from the equation y = x3+1 to the equation x = y3+1. Thisis contrary to the approach of our text book, the wording of the question, andthe approach I took in lecture. After a student writes this the reader cannotknow whether dy/dx means f0(x) or g0(x). I deducted three points (I thinkit is a serious error) and wrote: “Do not write both equations y = x3+ 1and x = y3+ 1 in the same problem. y 6= (y3+ 1)3+ 1.” Anticipating ageneral riot, I only deducted one point from the actual score I wrote on thefirst page of the exam. It was also common for students to write things likeg0(y) = (x−1)−2/3/3 rather than g0(y) = (y−1)−2/3/3 or g0(x) = (x−1)−2/3/3indicating that they don’t understand functional notation.IV. (30 points.) State and prove the formula for the derivative of theproduct of two functions. In your proof you may use (without proof) thelimit laws, the theorem that a differentiable function is continuous, and highschool algebra.1Answer: Suppose that f(x) = u(x)v(x) for all x where u and v are differ-entiable. Thenf0(a) = limx→af(x) − f(a)x − a(definition)= limx→au(x) · v(x) − u(a) · v(a)x − a(hypothesis)= limx→aµµu(x) − u(a)x − a¶· v(a) + u(x) ·µv(x) − v(a)x − a¶¶(hsa)=µlimx→au(x) − u(a)x − a¶· v(a) + u(a) ·µlimx→av(x) − v(a)x − a¶(limit laws)= u0(a)v(a) + v0(a)u(a) (definition)1This is one of five proofs the students were told to prepare.4(In the fourth step the theorem that a differentiable function is continuousis also used.)Grader’s commentsI would say half of the students got the problem essentially correct, eventhough it was clear that 90% of the time they just memorized the proof, asthey wrote down exactly what was on the handout. For the rest, some didn’tbother with the proof and most of them tried to reproduce the proof as wellas they could remember, and the result was typically having an intermediatestep that was completely inconsistent with the rest of the steps, or somethinglike starting to prove the product rule and by some mysterious algebra endingup with the quotient rule, or the addition rule. Nevertheless, there wereexceptions to this, and students that seemed to understand what they weredoing, and why.As to the grading, I took off three points if the hypothesis of the