Answers to Calculus Review, Differentiation. f() a) Use the definition of the derivative to calculate the derivative of f. First, you need the definition! I ll write it here in terms of Δ, although we ve also used h. f ( + ) f f lim 0 It s important to substitute + correctly. After that, it s a bunch of algebra. 4 4 f ( + ) f f lim lim +. Now that s ugly, but it s just a comple 0 0 fraction, and you can invariably clean those up by multiplying by the common denominator of the little fractions in both numerator and denominator. 4 4 4 ( + ) 4 ( + ) 4 4 lim + + lim + + lim 0 ( + ) 0 0 ( + ). Then distribute 4 4( + ) 4 4 4 4 4 lim lim lim lim. Now that the 0 ( + ) 0 ( + ) 0 ( + ) 0 ( + ) term is out of the denominator, there should be no trouble with letting equal zero. Here goes. 4 4 4 4 lim f 0 ( + ) ( + 0) Note that 4 d 4. Then [ 4 ] 4 4. It s a whole lot easier. (Remember now why you loved the power rule at first sight?) But that process doesn t use the definition of derivative, and would therefore not be a correct answer to the question I asked. The question of how complicated I m likely to make something like this on the test is completely legitimate. The answer is, not very. Which is to say that I m far more likely to ask you how to set up such a limit than I am to have you do all of the tedious algebra involved in evaluating it. I don t think that this algebra would be the best use of the time I have to evaluate your understanding of differentiation. But how to use the definition with any function I might throw at you that appeals to me. b) Write an equation of the tangent line to the graph of f at the point (, ). To get a tangent line, I need (or, rather, you need) a slope and a point. Hey, look, there s a point right in the question! To find the slope of a tangent line, a/k/a the slope of a curve at a point, you use the derivative. Therefore slope f ( ) And with that information, I can write an equation for the tangent. I ll use point-slope, partly because I m fond of it, mostly because I m lazy, and coincidentally because it will make the net part of the question really easy. y + ( + ) c) Write an equation of the normal line to the graph for f at the point (, ). The slope of the normal is, because the normal is perpendicular to the tangent. y + ( + ). a) DNE; there s a sharp corner at a. b) 0; the tangent line at b is horizontal c) negative; at c, the graph slopes down d) 0; at d, there s a horizontal tangent e) positive; at e (on the ais), k() slopes up f) DNE; the function isn t continuous at f g) DNE; there s a vertical tangent, and the slope of a vertical line is undefined.
Now for consecutive differentiate questions.. y + First, rewrite the radical as a power, then use the product rule. y + ( + ) / Using the product rule / / y ( + ) + ( + ) ( ) / / + + + That s okay, and you could stop there, but I m going to do the fancy algebra thing and factor out both the lowest power of ( + ), which is the power, and the constant coefficient,, because it s sort of ugly. y / ( + ) + ( + ) Where did the in the parentheses come from? If I were to multiply this back through, I d need to get one as the coefficient of the second part. and. Okay, but how did you get just plain ( + ) with it? It s the same thing, really: ( + ) ( + ) -/ gives ( + ) /, which was what I started with. More mathematically, factoring is dividing by the factor. If I divide ( + ) / by ( + ) -/, I subtract the eponents, and. Time for more simplifying. y + / ( + + ) / ( ) ( ) + 4 + 4 + 4 + y + + / Remember that in a multiple choice question, you have to be able to match the format of the answer (or at least figure out which answer matches yours), so occasionally doing the algebra is good practice. If you find the question and answer format here a little strange, you should know that it s how Plato wrote. And he s a famous philosopher. He wrote all of his stuff as dialogues between Socrates (his teacher) and an endless supply of people with questions. So there. And the cool looking symbol I used for the footnote is called a double-dagger. 4. f() ( + 5) 7 Chain, chain, chain, f () 7( + 5) 6 4 f () 8 ( + 5) 6 That was nice. 5. f() ( + )( + ) I have a couple of choices here. I could use the product rule, or I could multiply the binomials (you know, FOIL them) and just use the power rule. I m going with FOIL. f() 5 + + + Then f () 5 4 + + 6. 6. y + I could use the power rule right away, and the chain rule would have me use the quotient rule to get the derivative of the inside. I could do that, but I won t. Why not? Because the power just means that I need to (all together now ) Flip That Sucker Over! Watch. + y + Isn t that better? No chain rule, just a quotient. However, because I m the math teacher, I m going to be even sneakier and break that fraction into two pieces. If I divide each of the terms in the numerator by, I wind up with y +, and all I ll need is the power rule. Any of those three is correct. So you can see that it s the same, I ll work out the chainquotient combo, too. ( ) ( + ) + ( + ) + + + + And there you go.
7. r() + 5 + 5 I made it into powers to use the power rule. r () 6 + 0 0 r () 6 + And I put it back into fractions to match the original format. 8. y You could do this problem with the quotient rule, but it s overkill, because π is just a constant. If you rewrite y as a power, you get a simpler function to differentiate. y π π 9. y π The similarity to the previous question is intentional. The π is still a constant, and you still don t need the quotient rule. If, instead of writing it as a division, I write it as a multiplication, it s really simple. y 0. y + csc + cot This one is very straightforward, provided you still remember the derivatives of all si trig functions. y 0 + csc cot csc y csc cot csc, or, factoring, y csc ( cot + csc ). g() tan (5 sin ) Use the chain rule. g () sec (5 sin ) cos Did you catch that etra on the end? There s a second inside in the sine function; I multiplied by its derivative, too. g () cos sec (5 sin ). y 5 cot ( ) I hear Aretha calling out to me! y 5 csc ( ) 0 csc Some of you might be wondering if you can turn the cosecant of that ugly fraction into the sine of its reciprocal. The answer is no. The argument of cosecant is an angle, not a ratio, and the reciprocal of an angle doesn t make much sense mathematically.. y sin ( + y) Since you can t isolate y, use implicit differentiation. I m going to use for the derivative of y. I could use y. It s all the same. cos ( + y) + Distribute. cos ( + y) + cos ( + y) Collect terms on the left. cos ( + y) cos( + y) Factor out that. ( cos( + y) ) cos( + y) Notice the there; it matters. Then divide. cos( + y) cos( + y) 4. g() (6 ) /4 The quotient rule is overkill when the numerator is a constant, since its derivative is 0. It s much more sensible to write those as powers instead. g () (6 ) 5/4 g ()
5. y I could solve for y, but it would be harder to differentiate, and it would involve a ±, and I just don t need that. So it s implicit for me y y 6. y + y 7 Clearly this one is implicit, too. Differentiating, y + y 0 + y y (y ) y If you had moved the terms to the right side, instead, you d have ended up with y, which is the same thing. Notice the y symmetry, too. 7. y ( + ) / Using the product rule, 8. ( + ) / + ( + ) / + + + And that s fine. But here s some fancy factoring like I did back in problem. ( + ) / ( + ( + ) ) U-G-L-Y, you ain t got no alibi uh, never mind. It s just the chain rule, and it s actually much prettier with powers than it was with roots. / / 4 4 ( / ) 9. y 4 y Implicit differentiation: 4y y ( ) Remember that it s only the terms with y in them that need the on the end. Collect terms on one side. 4y y Factor out the. (4y y) Divide. 4y y y y y y 0. cos y sin y 0, which means (cos y) sin y 0. Differentiating with the product rule, (cos y) sin y + (cos y) cos y 0 cos y sin y + cos y cos y 0 cos y cos y sin y cos y + cos y ( cos y sin y + cos y) cos y cos y sin y + cos y Yes, it would be nice to cancel the s, but I can t do it. It s not a factor of the entire denominator.. f() ln Product rule f () + ln + ln 4
4 +. y ln This is not the prettiest question. You could certainly work it as it stands, but the fraction and radical are not a great combination. So let s Harness the Power of the Logarithm. 4 + y ln ln(4 + ) ln That s better! 4 + 4 +, or, if you re obsessive, (4 + ) 4 (4 + ) (4 + ) If you re wondering why I didn t do that part where you multiply by y at the end, it s because this isn t really the process called logarithmic differentiation. That s when you have something so hideous that you actually take the logarithms of both sides of the original formula for y. In that case, differentiating the left side gives an etra y. It s not on the test.. y ln Yes, I could take the derivative of this as-is, but I m way too lazy for that. y ln ln ( ) ln. Ah, yes! 4. g() lnsec + tan There are no fancy tricks to make this one better. It s just the chain rule. g () sec + tan (sec tan + sec ) sec tan + sec sec + tan That result, on the other hand, cleans up quite nicely. Actually, this isn t meant to be a footnote; I m marking the phrase as my registered trademark. I m registering it with you, right now. sec (tan + sec ) sec + tan sec 5. y ln (sin ) cos cos cot sin sin Nice. 6. y (ln ) 5 I see you deriving in class during calculus, and you ve got f (u) 4 5(ln ) 5(ln ) 7. g() arctan g () + (You memorized this one, right? You really need to memorize all of those. You ve been warned.) 8. h() arccsc (e ) Use the chain rule. The derivative of arccsc is, if you ve forgotten. h e e e e e 4 e e Note that because e is always positive, I was able to eliminate the absolute value and reduce. 9. y e e e + e It s a quotient rule. It will turn out fairly well, but the intermediate steps are not so nice. ( e + e )( e + e ) ( e e )( e e ) e + e The numerator simplifies. Watch. ( e + e e + e ) e e e + e ( e + e ) e + e e + e e + e e e ( e + e ) 5
4e e 4e ( e + e ) ( e + e ) 0 4 Did you catch that slick add-theeponents thing? Niiiiice. sin e + e 0. f () loga a That simplifies. Logarithms and eponential functions are inverses of each other. f () log a a sin sin f () cos. y e Just a nice simple chain rule. e / e. y Product rule. Remember how to take the derivative of an eponential function with a base other than e? It s just like e, but multiplied by the natural logarithm of the base. y ln + y ln +. g() ln 0 HtPotL! g() ln 0 ln 0 ln g () 4. y ln 5 More harnessing. Let s go. y ln 5 Now take derivatives. 5 + 5 ln ln ( 5 ) 5. y log4(4 + ln 4) The derivative of a log of a base other than e has the natural log of the base in the denominator. ln 4 4 ln 4 ln 4 ( + ) 4 + ln 4 6. To find a numerical derivative value without first finding the analytic (algebraic) formula for f () first, you use the Numerical Derivative at a Point command on your calculator. These instructions assume you re using a TI-nspire. It s easier to get the synta right if you choose it from the Calculus menu. Some screen shots: You type the value of in the Value: bo. (I know, helpful name, right?) Then Enter. And the screen will look like this: Then type the function in the bo, and Enter again. If you didn t get this answer, you re probably not in radian mode. TZ (Yes, that s how they punctuate it. I m looking at the logo now. The TI part should really be small caps, but I m not up for that level of verisimilitude at the moment.) TZ If you re using a TI-8 instead, the command you want is nderiv( in the Math menu. The synta is nderiv(function, variable, value), so in this case it looks like 6
7. Let f() 0, f () 6, g(), and g (). Find stuff, if possible. a) h () if h() f()g() Use the product rule. h () f() g () + g() f (). So h () 0 + 6 6 b) j (), if j() g(f()) This time it s the chain rule. j () g (f()) f () j () g (f()) 6 g (0) 6 6g (0) Unfortunately, I don t have any information about g (0), so I m stuck there. g c) k (), if k h Lo-d-hi! I mean, quotient rule! h() g () g() h () k () (( h()) Ummm h() isn t in the list. Waitwaitwait! That function was in part (a)! So h() f()g() 0 0, and h () 6, which was what I found in (a). Back to k. 0 6 6 k () 0 0 So k () doesn t eist. But k does have a vertical tangent when, so I did get some information out of this. (BTW, did you like my simulation of ecitement in the previous paragraph?) d) m (), if m() f() g() If you need a name for it, this is the constant multiple rule. m () f () g () m () f () g () 6 7 8. A particle starts at time t 0 and moves along the -ais so that its position at any time t 0 is given by (t) (t ) (t ). a) Find the displacement of the particle from t to t 5. Displacement is change in position. Position at time is () ( ) ( ). Position at time 5 is (5) (5 ) ( 5 ) 4 7 448. So displacement is 448 447. No units in the problem means no units in the answer. b) Find the average velocity of the particle from t to t 5. Average velocity is change in position divided by change in time. Change in position is the displacement, which I just found, and change in time is. So 447 49. c) Find a formula for the instantaneous velocity of the particle at any time t 0. v(t) (t), which means use of the product rule. v(t) (t ) + (t ) (t ) (t ) (t + 6t 9) (t ) (8t ) Yes, I factored in the second line. You could have left it as v(t) (t ) + (t )(t ), but I looked ahead, and I m going to need a second derivative. The nicer it looks at this step, the fewer opportunities I have to make mistakes later. d) For what values of t is the velocity of the particle less than zero? v(t) < 0, so (t ) (8t ) < 0 The zeros of (t ) (8t ) are t and t. The simplest way to solve the inequality is to use a number line and test points. Do you remember this from back in algebra? The domain starts at 0, and I ll mark circles at the zeros of velocity, since I m only looking for times when velocity is strictly negative. I ll look at what happens to the sign of v at a value in each of the three intervals I get. Since the first part of v is squared, it won t affect the sign, and I only have to consider the sign of 8t. For values between 0 and, 8t <, and v will be negative, which is why that s shaded on the number line. For values between and, 8t <, so 8 8 7
v is still negative, and that section is shaded. Finally, for values on the right end, 8t >, and v is positive. So v(t) < 0 for t [0, ) (, ). e) Find the acceleration of the particle at any time t 0. a(t) v (t). Using the product rule, a(t) (t ) 8 + (8t ) (t ) (t )(4t 4 + 8t ) (t )(t 5) 6(t )(4t 5) No, you didn t have to make this as pretty as I did. f) Find the value of t when the particle is moving and the acceleration is zero. First, find the times when acceleration is zero. a(t) 6(t )(4t 5) 0 t or t.5 But now I need to see if the particle is moving. That means its velocity has to be nonzero. v(t) (t ) (8t ), so v() ( ) (8 ) 0. v(.5) (.5 ) (8.5 ) 0.065 0.065. Therefore t.5 is the only time that satisfies the requirement. True or false? Justify. 9. If f (7) 5, then f is continuous at 7. True. Differentiability implies continuity. 40. If f (7) 5, then f is decreasing at 7. False. The slope is positive, so the function is sloped up, and thus increasing. 4. If f (7) 5, then y 5 5( 7) is tangent to f at the point where 7. False. The slope is correct, as is the value of at the point, but there is no reason to believe that the y-value there is 5. f (5 + ) f (5) 4. If f (7) 5, then lim 7. 0 False. The 7 and the 5 are backwards. That limit really says that f (5) 7. f f (7) 4. If f (7) 5, then lim 5. 7 7 True. It s an alternative definition of the derivative at 7. 44. If f (7) 5, then ( f ) (7) 5. False. It s definitely correct that the slopes of inverse functions are reciprocals of each other, but the -value on f isn t (necessarily) the same as the -value on f. The formula for calculating the derivative of an inverse function is ( f ), so to find f f ( f ) (7), I d need f (7). It s not here. 45. The balloon is rising at m/s, which gives, and the distance from A to C is fied at 00 m. a) Find the rate of change in at the instant when y 50. I ll need an equation to relate and y. I m thinking of going with the Pythagorean theorem. 00 + y Differentiating,. I m given y and, and I m to find, but I still need a value for. And here comes Pythagoras, my ancient Greek math superhero, to save the day! When y 50, 00 + 50, so. Substituting, 50 50 50 500 50 5.4 m/s 5 b) Find the rate of change in the area of the triangle when y 50. Hmmm, a formula for the area of a triangle! A bh 50y Notice that it s okay for me to substitute 00 for the base since it s not changing. I can t substitute a value for y yet, because that is changing. Differentiate. 50 50 m /s 8
This result is independent of the value of y; it only involves the rate of change of y, which was given to be a constant m/s. Nice, huh? c) Find the rate of change in θ when y 50. I need an equation to relate θ and y. I could use any of the trig functions, but it will be easiest if I use one where one side is constant. I pick tangent. tan θ Differentiating, sec θ. You may recall in part (a) that when y 50,. So sec θ. Then sec θ becomes area. It s A 4πr. So take some derivatives. da dr 8π r That s nice and all, but I only know dr when r. I hope that s the same here. I m given the volume, so I should be able to find the radius for certain. π 4 π r π 4πr 8 r r. Ecellent! Therefore I can use the value from part (a), dr 4.5 6π. da dr Now 8π r 8 π 4.5 6π 4.5 in /min. Ahhh, success...8 /s 0.04 radians/s 46. At long last. Air is being pumped into a spherical balloon at 4.5 in per minute. That s dv, and it s positive, since air is being pumped in. a) Find the rate of change of the radius when the radius is inches. The problem gives the formula for the 4 volume of a sphere as V π r, so I ll differentiate that. dv dr 4π r Substituting what I know gives 4.5 4π dr, and thus dr 4.5 0.090 in/min. That s 6π very slow. In fact, 4.5 in per minute is really slow, too. b) Find the rate of change of the surface area at the time when the volume is 6π in. So I need a formula for the surface 9