C6-2 Differentiation 3

Transcription

1 chain, product and quotient rules C6- Differentiation Pre-requisites: C6- Estimate Time: 8 hours Summary Learn Solve Revise Answers Summary The chain rule is used to differentiate a function of a function. The outside function is differentiated with respect to the inside function as if it is, but then the result is multiplied by the derivative of the inside function. If y = y(u()), then dy d = dy du du d. The product rule is used where a function can be written as one function multiplied by another function. If y = uv, then y = u v + uv, where u is the derivative of u etc. The quotient rule is used where a function consists of one function divided by another function. If y = u u v uv, then y = v v Some functions require more than one application of these rules. Learn a n, sin, cos, a and ln are the basic functions which you know how to differentiate. You also know how to differentiate sums of these and multiples of them. You don t yet know how to differentiate more comple functions like e sin, sin or ln ( + ). All of these types of functions, and in fact just about any function you will ever meet, can be differentiated using what you already know plus three rules the chain rule, the product rule and the quotient rule. Once you can use these three rules, you will be able to differentiate any function you are ever likely to meet. The Chain Rule Function of a Function We use the chain rule to differentiate a function of a function. is a function of. Think of a function as something we do to. In this case we squared it. In the same way, sin is a function of something we did to in this case we took the sine of it. Black Star Maths C6- Differentiation Page

2 But, as is a function of, sin is a function of a function of. Here, the sine is the outside function done last) and squaring is the inside function (done first). To differentiate sin, first let the inside function (the squaring) be u, so u =. Then let the outside function (the sine) be y, so y = sin u. dy du Then we find and. du d dy dy du Then =. We can see this to be true just by cancelling the dus. d du d A good way to lay this out is like this: y = sin y = sin u u = dy = cos u du dy dy du = d du d = cos u = cos du = d = cos Go over this again until it makes complete sense. Here s another eample. Let s find d (e + cos ) 4. d y = (e + cos ) 4 y = u 4 u = e + cos dy = 4u du dy dy du = d du d = 4u (e sin ) du = e sin d = 4(e + cos ) (e sin ) Black Star Maths C6- Differentiation Page

3 P Use the chain rule with the layout above to differentiate the following. Don t forget to get the functions into differentiable form first by epressing fractions as negative powers and roots as fractional powers. (a) y = sin (b) y = sin ( + 5) (c) h = (sin t + ) 5 (d) y = ( + 5) 8 (e) h = 4 ( ) (f) h = (g) y = (ln ) (h) h = cos () (i) r = ln (t t ) (j) y = cos (t ) (k) P = e sin (l) s = (m) y sin (n) t = cos (e) (o) A = 4 e 5 Once you have had a bit of practice with this, it is possible (and in fact preferable) to do the differentiations without writing down the working steps. To do this, we think about it just slightly differently like this: Suppose we need to differentiate sin ( + 7). Up to now, we would think of this as y = sin u and u = + 7. dy du And we would have written du d Now, instead, we look at the outside function and think of this as the sine of something. Then we differentiate the sine of something, as if the something were, to give the cosine of the same something, which is cos ( + 7). Now, if the something had just been, then we would be finished. But, because the something is not but rather a function of, we then have to multiply by the derivative of the something, in this case + 7. And we get cos ( + 7) ( + 7). dy du Note that the + 7 is the u, the cos ( + 7) part is the and the ( + 7) part is the. So du d we have done the same thing as before, but without writing out any working steps. As another eample, let s differentiate ( + ) 9. The outside function is raising to the power of 9. So the epression is something to the power of 9. So we differentiate to get 9 something to the power of 8, i.e. 9( + ) 8. Then we multiply by the derivative of the something to get 9( + ) 8 ( + 6). P Use the chain rule to differentiate the following without writing any working. (a) y = ( + ) 6 (b) y = (sin ) (c) h = (cos r + r ) 5 Black Star Maths C6- Differentiation Page

4 (d) y = cos (4 + ) (e) h = (sin ) (f) h = 4 t (g) y = sin (ln ) (h) h = cos (5 ) (i) r = (t t ) 5 (j) y = 7 sin (t 4 ) (k) P = sin (l) s = (m) y 4 cos (n) t = sin (e) (o) A = (p) = ln (sin t) (q) p = sin cos (r) y = (s) f = sin ( ) (t) t = sin s (cos s) (u) w = e ln 5 (cost) Some of the derivatives can be simplified and it is good style to simplify them where possible. For eample, P(s) gives cos ( ). This can be simplified to 6 cos ( ). You should simplify answers where possible from here on and answers will be given in simplified form. P Differentiate the following without writing any working, but simplifying your answer where possible. Hint: Rewrite sin as (sin ) etc. before differentiating, but put them back into the sin form when simplifying. (a) y = sin 4 (b) y = sin ( + ) (c) y = (cos t + t) (d) y = ln (e) h = cos ( + ) (f) r = ln (a + sin a) (g) y = 4sin (t ) (h) P = (7 ) (i) s = 6 e (j) y 7 cos (k) t = 4 + sin (e) (l) A = (m) = ln (t + cos t) (n) p = n n (o) y = 4 (p) f = sin (4 ) (q) t = cos s cos s (r) b = (cost) 7 The Chain Rule Function of a Function of a Function Suppose we need to differentiate sin e. is a function; e is a function of a function; and sin e is a function of a function of a function. Differentiating a function of a function of a function is really no harder than differentiating a function of a function. Black Star Maths C6- Differentiation Page 4

5 Again we look at the outside function, in this case sine of something. The derivative is cos of something, in this case cos e. Then, because the something isn t just, we have to differentiate by the derivative of the something. Now, in this case, the something is a function of a function and so requires the chain rule. Its derivative is e. So the whole derivative is cos e e, which should be simplified to e cos e. A function of a function of a function of a function can be differentiated in the same way. Just start with the outside function. Then the derivative of the something will be the derivative of a function of a function of a function. You might be able to see why this rule is called the chain rule. P4 Differentiate the following. Don t forget to simplify where appropriate. (a) y = cos e 4 (b) y = sin e 5 (c) y = ln (cos t) (d) y = sin (ln 6) (e) h = e cos (f) r = ln (sin a ) (g) w = sin (ln ) (h) P = (cos 4 ) (i) h = e 4cos e sin (j) r = sin (e ln ) (k) y = ln (cos ( )) sin (+) You should get to the point of being able to use the chain rule fairly automatically without having to puzzle over it or to remind yourself of what to do. Practise until you get to that point. The Product Rule The product rule can be used for any function which can be written as one epression multiplied by another epression. Consider y = 4 sin. This is not a function of a function; it is two functions multiplied together, i.e. a product of two functions. The way to tell if a function is a product and therefore if we can use the product rule is to see if it can be split with a vertical line into two parts which are multiplied. 4 sin can be split thus: 4 sin and written (4) (sin ). sin (4 + ) cannot be. Neither can (ln ). They are both functions of functions rather than products of functions and so require the chain rule. Black Star Maths C6- Differentiation Page 5

6 The product rule is easy. Again, we call the function y. We also call the two parts of the product u and v. du dv We call the derivatives u and v respectively, i.e. u = and v =. The u and v notation d d makes it quicker to write out as well as easier to remember. In the case of 4 sin y = 4 sin u = 4 v = sin. u = 4 v = cos y = u v + uv = 4 sin + 4 cos It is recommended that this layout be used as it provides cues as to what to do net. P5 Use the product rule to differentiate the following, laying out the working as shown above. (a) = sin (b) y = e sin (c) y = ( + 5) cos (d) y = ln (e) h = sin cos (f) r = ln (g) y = 4e t sin t (h) P = ( 5)(4 ) (i) s = 6e ( ) (j) k = ln (k) t = 4 cos (l) A = As with the chain rule, once you ve had a bit of practice laying out the working, it is good to learn to do it without the working. It can be helpful to think of it as writing out uv twice, first time with u differentiated, second time with v differentiated. P6 Use the product rule to differentiate the following without writing any working. (a) y = sin (b) v = (5 + ) e (c) P = (cos t + t) sin t (d) y = e sin (e) h = ( 5) ln (f) r = (cos w + sin w) sin w (g) y =.05 (h) P = (7 ) cos (i) s = 4 ( ) e (j) f = e (k) t = 4 sin (l) A = + sin (m) = t ln t e t (n) p = n cos n (o) y = 5 sin cos (p) f = e sin (q) t = cos s s cos s (r) b = (r 4r)(cos r sin r) Black Star Maths C6- Differentiation Page 6

7 If you meet an epression which is a product of functions, the product rule can be adapted to y = u vw + uv w + uvw. This time we write out uvw times, first time with the u differentiated, second time with the v differentiated and third time with the w differentiated. You can probably etrapolate to the formula for a product of 4 functions, although you might run out of letters. P7 Use the product rule to differentiate the following. (a) y = e sin (b) y = ( + ) cos (c) y = ( + 5) e ln The Quotient Rule The quotient rule can be used for any function which can be written as one epression divided by another epression. Consider y = e We let the top epression (the numerator) be u and the bottom epression (the denominator) be v. Again, we let their derivatives be u and v. We write: y = e u = e v = + u = e v = + y = = u' v uv' v e ( ) e ( ) ( ) = ( ( ) e ) (Note that the top of y = u' v uv' v is the same as for the product rule ecept for the minus.) Once again, this layout is recommended while you get used to the rule. Once you are confident, you should be able to use the quotient rule without writing down any working. Black Star Maths C6- Differentiation Page 7

8 P8 Use the quotient rule to differentiate the following. (a) y = + (d) y = e sin (g) y = cos (j) y = 4 ln (m) y = (b) y = + (e) y = cos cos (h) y = sin ln (k) y = ln (n) y = + sin (c) y = ln (f) y = + (i) y = tan (l) y = e ln ln (o) y = +sin Mied The questions in the net eercise require the chain rule, product rule or quotient rule. They will give you practice in choosing as well as more practice in using the rules. P9 P0 Use the chain rule, product rule or quotient rule to differentiate the following. (a) y = sin 4 (b) y = sin (c) s = 6 e ln (d) y = (e) h = e cos (f) r = ln (a + sin a) (g) y = 4 sin (t ) (h) P = 9 + (7 ) (i) y = (t+ sin t) (j) 7 y (k) t = 4 + sin (e ) (l) A = 8c sin c cos (m) = ln (t + cos t) (n) p = n n (o) y = (p) f = e (q) t = sin s + sin s (r) b = 4 cost Use the chain rule, product rule or quotient rule to differentiate the following. (a) y = cos ( + e ) (b) z = sin 6 (c) m = 4 ln sin (d) y = (e) h = e (f) r = 4e 5s 4 (g) y = 4t 6 sin t (h) P = 5 (i) y = (n ln n) (j) 7 y (k) t = 4e + sin (l) A = ln c sin c cos e t Black Star Maths C6- Differentiation Page 8

9 (m) = e t sin t (n) p = 4 sin r (o) y = (p) f = e (q) a = cos + cos (r) = 5 t sin t Rules in Combination Some functions require more than one application of the chain, product and quotient rules. y = sin is an eample. y = (ln ( + )) is another. To differentiate these, we look at the overall structure: if a vertical line can separate it into two functions multiplied together, then start with the product rule; if it is a function over another function, then start with the quotient rule; otherwise, if it is a function of a function, start with the chain rule Eample Taking y = sin, this is a product, so we think u v + uv and write sin + the derivative of sin. But, of course, differentiating sin requires the chain rule, giving us cos. So we get sin + cos, which can be simplified to sin + cos. Eample Taking y = (sin ( +)), this is a function of a function, so we use the chain rule to get (sin ( +)) the derivative of sin ( +). But differentiating sin ( +) requires us to use the chain rule again to get cos ( +) ( + ). So we get (sin ( +)) cos ( +) ( + ), which can be simplified to (6 + 6) sin ( +) cos ( +). Basically, you just follow your nose and it should come out right. The following questions will give you the practice you need. P Use the chain rule, product rule and/or quotient rule to differentiate the following. (a) y = sin (b) y = ln ( sin ) (c) s = 6 e (d) y = ln (e) h = e+ sin (f) r = e sin (g) y = sin (t e t ) (h) P = (sin ) 5 (i) y = (t cos t) e t Black Star Maths C6- Differentiation Page 9

10 (j) y (k) t = 4 e sin sin (l) A = b sin (6b + b) (m) = ln (t 4 sin t ) (n) p = n sin n (o) y = e cos P (p) f = + 4 e 6 (q) t = [ln (sin s)] (r) b = Differentiate the following. (a) p = sin 5 cost 5 e t cos (b) b = (c) y = (cos t + t) (d) y = ln (e) h = cos ( + ) (f) r = ln (a + sin a ) (g) y = 4t sin (t ) (h) P = cos (7 ) (i) s = ln ( cos ) (j) y (k) h = sin + tan (l) A = cos (m) z = t t t 4 (n) g = (p) p = n n n (q) y = b (s) c = 5 (cosb) cos (o) = 4 ln (t cos t) sin 4 (t) f = cos sin (4 ) (r) t = cos s sin s (u) y = cos tan And a few word problems: P The tide height in metres at Wilhelm Bay last Saturday was given by h =.8 cos ½(t ) +., where t is the time in hours since midnight. How fast was the tide rising or falling at 0 am? P4 The population of Nobbsville is given by p = e 0.005t, where t is the number of years since 990. How fast was the population growing in 07? P5 Julie s height in centimetres between the ages of 4 and was given by h = 65. a, where a is her age in years. Her mass in kilograms is given by m = 0.004a h. At what rate was her mass increasing on her 9 th birthday. P6 The average price of houses in Blurbledale in thousands of dollars between 994 and 008 was given by p = t 4t+0. How fast were prices rising or falling in 5+ t 00? Black Star Maths C6- Differentiation Page 0

11 Solve S What you have learnt about differentiation should enable you to differentiate just about any function you will meet, however comple. Differentiate ln sin ( e cos ) S Differentiating y = is a bit of a challenge. It can be done with what you know about differentiation and some manipulation using what you know about logs. Try it. S A pendulum swings such that its displacement,, to the right of its equilibrium position is given by = 8 e 0.04t cos t, where is in metres and t is the time in seconds. The eponential factor is there because the amplitude of the swings decreases eponentially with time. Find an epression for the horizontal velocity and hence find the velocity the fourth time it is at the equilibrium position. Revision Set R R Differentiate: Revise (a) y = ( + 7) 8 (b) y = e (c) y = (d) p = r ln r (g) y = cos ( e ) + (e) m = sin cos (h) y = ( + 7) sin +5 (f) y = sin ( + e ) Arthur walks km east from camp, then 4 km north. He is then 5 km from camp. After that, he continues to walk north at km/h. At what rate is his distance from camp increasing after he has walked a further km, i.e. when he is 7 km further north than camp. Revision Set R R Differentiate: (a) y = cos (b) y = tan (c) h = 4e 0.t + t (d) s = sin (g) y = ln ( sin 5) (e) y = sin ( ln ) (h) y = sin( ) 4 (f) y = cos ln A weight hanging on the end of a string is swinging, but the amplitude of the swings is slowly decreasing. It s sideways displacement t seconds after starting is given by = e 0.0t cos t. What is the horizontal component of its velocity when t =? Revision Set R Differentiate: (a) y = 7 0 cos (b) A = e + sin (c) y = sin Black Star Maths C6- Differentiation Page

12 R 4 7 (d) y = 4 (e) h = 4 ln cos t (f) k = e cos (g) y = sin cos ( + 4) t (h) b = ln cost If t is the time in days since the start of April, the probability that it will have rained by time t is given by p = ¼ln (t + ) cos 0.005t. At what rate is the probability increasing when t = 0? Answers P (a) s = cos (b) y = (6 + 5)cos ( + 5) (c) y = 5 cos t(sin t + ) 4 (d) y = 8( + 5) 7 (6+5) (e) h = 4( ) ( ) 5 (f) h = (g) y = (ln ) (h) h = sin (i) r = t t t (j) y = sin (t ) (k) P = cos e sin (l) s = 8 e 4 cos (m) y = sin (n) t = e sin (e ) (o) A = ln 5 5 P (a) y = 6( + ) 5 (4 + ) (b) y = (sin ) cos (c) h = 5(cos r + r ) 4 (r sin r) (d) y = (8 + ) sin (4 + ) (e) h = cos(ln ) (g) y = cos 4 (sin ) (f) h = t 4 t (h) h = 5 sin (5 ) (i) r = 5(t t ) 4 ( t) (j) y = 8 t cos (t 4 ) (k) P = sin ln cos (l) s = e (m) 4sin ' (n) t = e cos (e ) (o) A = cos y cos t (p) = sin t (s) f = 6 cos ( ) (q) p = cos sin sin cos (r) y = ln (t) t = s cos s 5 sin t cos s sin (u) w = (cos t) 6 ln P (a) y = 4 cos 4 (b) y = 4 cos ( + ) (c) y = (cos t + t) ( sin t) (d) y = (e) h = sin ( + ) +cos a (f) r = a+sin a (g) y = 4t cos (t ) (h) P = (7 ) 0 (54 ) (i) s = 6 e 7sin (j) y' (k) t = 4 + e cos cos (e ) (l) A = 9 ln t sin t +n 4 (m) = t (n) p = (o) y = +cos t n 4 +n +n (4 ) Black Star Maths C6- Differentiation Page

13 4 sin t (p) f = (4 ) cos (4 ) (q) t = s sin s + sin s cos s (r) b = 8 (cost) P4 (a) y = 4e 4 sin e 4 (b) y = 5 4 e 5 cos e 5 (c) y = tan t cos (ln 6) (d) y = a (e) h = sin cos e cos (f) r = tan a (g) w = 4 ln cos (ln ) (h) P = (4sin 4 )(cos 4 ) 0 (i) s = 6sin cos e 4cos sin cos e sin (j) r = e ln cos e ln ln (k) y = cos(+) tan( ) sin (+) sin (+) P5 (a) = sin + cos (b) y = e (sin + cos ) (c) y = ( + 5) cos ( + 5) sin (d) y = ln + (e) h = cos sin (f) r = ln ln + (g) y = 4e t (sin t + cos t) (h) P = (i) s = (6 + )e ln + (j) k = (k) t = 4 sin 4 cos (l) A = ( + ln ) P6 (a) y = cos + sin (b) v = ( ) e (c) P = cos t + t cos t + sin t sin t (d) y = e (sin + cos ) (e) h = ( 5) ln + 5 (f) r = ( cos w sin w) sin w + (cos w + sin w) cos w (g) y =.05 ln (h) P = (4 ) cos + (7 ) sin (i) s = ( ) e (j) f = ( + 4) e (k) t = 4(sin + cos ) (l) A = + cos (m) = ln t + e t (n) p = cos n n sin n (o) y = 5 (cos sin ) n (p) f = e sin + e cos (q) t = (s )sin s cos s (r) b = (6r 4)(cos r sin r) (r 4r)(cos r + sin r) P7 (a) y = e sin + e sin e cos (b) y = ln ( + ) cos + ( + 4) cos ( + ) sin (c) y = ( ) e ln P8 (a) y = +6 (+) (b) y = ( )( + ) (+)( ) cos sin ( + ) (c) y = (d) y = ( )e (e) y = (ln cos + sin ) cos (g) y = cos (h) y = sin + ln (f) y = (+) (i) y = cos Black Star Maths C6- Differentiation Page

14 (j) y = 4(ln ) (ln ) (k) y = ln (ln ) ( ln )e (l) y = (ln ) (m) y = ( +)(8 6) (4 6) ( +) (n) y = +6 (+) + sin ln cos (o) y = (+sin ) P9 (a) y = 4 cos 4 (b) y = cos + sin (c) s = 6 e (d) y = ln +cos a 4 (e) h = e (cos sin ) (f) r = a+sin a cos t (g) y = 4t cos (t ) (h) P = (7 ) 0 (4 ) (i) y = (t+sin t) (j) 7cos 7sin ' (k) t = 4 + e cos cos (e ) (l) A = 8sin c + 8c cos c y t sin t (m) = t +cos t +n 4 (n) p = (n+n ) (o) y = (4 ) (p) f = ( + ) e (q) t = s cos s + sin s cos s (r) b = e t (sin t cost) cos t P0 (a) y = ( + e ) sin ( + e ) (b) z = 6 cos 6 c) m = ln (d) y = cos sin (e) h = e (f) r = 0 e 5s (g) y = 4t 6 cos t 4t 5 sin t (h) P = 4(5 ) (5 4 ) 4 n (i) y = n(n ln n) (j) ( 5 ) 4 cos 7 sin sin c ' (k) t = e + 6sin (l) A = ln c cos c + cos c y (m) = ( cos t) e t sin t cos r (n) p = 4 (o) y = 4 sin r 5(6 ) ( ) (p) f = ( + ) e (q) a = sin 6 cos sin (r) = t t ln sin t cost sin t sin + cos P (a) y = cos + sin (b) y = (c) s = (6 4 + ) e sin ln (d) y = 4 (e) h = e + (cos + sin ) (f) r = e (sin cos ) sin (g) y = e t (t + 6t) cos (t e t ) (i) y = et (t cos t sin t (t cos t) (h) P = 0(sin ) 4 (cos ) (j) y = sin cos sin (k) t = 4e ( cos + sin + sin ) Black Star Maths C6- Differentiation Page 4

15 (l) A = sin (6b + b) + b(b + ) cos (6b + b) (m) = (t5 cos t + 4t sin t ) +n cos n t 4 sin t (n) p = n+sin n cos +sin (o) y = e cos cos (p) f = e 6 (4 6) 6[ln (sin s)] (q) t = tan s (r) b = sin ln cos P (a) p = sin cos 5 (b) b = (c) y = (cos t + t) ( sin t) (d) y = ln + (e) h = 4 cos ( + ) + 4 sin ( + ) +a cos a (f) r = a+sin a 5 e t (0 cost cos t 4t sin t (g) y = 8t sin (t ) 4t cos (t ) (h) P = (4 8) cos (7 ) sin (7 ) cos sin (i) s = cos (k) h = cos + cos 4 4t t (m) z = (4t t) (n) g = (j) (l) A = cos ' cos y sin ( 9 ln ) cos sin 8 cos t 4t sin t (o) = cos t cos t 4n +5n 4 (8 4) cos 4sin (p) p = (n 4 +n 5 ) (q) y = (4 ) (r) t = sin s cos s cos s s sin s sin s 0 sin b b cosb (s) c = 6 (cosb) (t) f = sin sin (4 ) + (4 ) cos cos (4 ) (u) y = sin P P4 P5 P6 Rising at 0.6 m/h 8.5 people per year 75 kg/year (Julie was a whale.) Falling $800/year S y = sin ( e cos ) + ln (cos ( e cos ))( e cos sin e cos ) S y = ( + ln ) S = 8 e 0.04t (0.04 cos t + sin t);.84 m/s R (a) 8( + 7) 7 ( + 7) (b) ( + ) e (c) y = 4+5 ( +5) (d) r ln r + r (e) m = cos (f) ( + e ) cos ( + e ) (g) ( + ) e sin ( e ) + (h) ( + 7) sin + (6 + 4) sin cos R 0.99 km/h Black Star Maths C6- Differentiation Page 5

16 R (a) y = cos + sin (b) y = R sin cos (d) s = sin sin 5+5 cos 5 (g) y = sin 5.48 m/s in the positive direction cos cos ( ln ) (e) y = (c) h = 0.4e 0.t + (h) ( )cos( ) 4 ln sin( ) 4 R (a) y = 7 0 (ln 0 cos sin ) (b) A = R (d) y = 6 4 (4 4) (4 ) cos (f) y = sin ln ( ) e + cos (c) y = (e) h = 4 tan t (f) k = e (cos sin ) (g) y = cos cos ( + 4) [cos ( + 4) + 6 sin ( + 4)] (h) b = 0.0/day cos sin cost t sin t t cost Black Star Maths C6- Differentiation Page 6