Questions on Differentiation
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1 Questions on Differentiation In Questions 3-8 I give some of the eamples of functions that have ha to be ifferentiate in the previous few eams for this course. 3 (2003 Differentiate i ln ( +, for >, twice, : The Sum Rule gives (ln ( + ln ( + +, 2 (ln ( + 2 ( + 2. ii 2 + tan 2, when (2k + π/2 for any k Z. : ( 2 + tan (tan tan ( 2 + tan tan cos 2. (Sum Rule ( tan + tan tan by Prouct Rule When ealing with trigonometric functions it can be har to know how to leave the answer because of the trigonometric ientities. You might think it nicer to epress the answer as (tan ( + tan 2 because no fractions are involve. iii tan, when (2k + π/2 for any k Z. : ( tan ( tan + tan, tan + cos 2. Again you might epress this as tan + ( + tan 2. (Prouct Rule
2 iv e sin, 9 times. : Let f ( e sin. Then f ( ( e sin + e cos, f (2 ( 2e cos, f (3 ( 2e cos + 2e sin, f (4 ( 4e sin, f (5 ( 4e sin 4e cos, f (6 ( 8e cos, f (7 ( 8e cos 8e sin, f (8 ( 6e sin, f (9 ( 6e sin + 6e cos. 4 (2004 Differentiate cos 2, 8 times, : Let g ( cos 2. Then g ( ( 2 cos sin sin 2, g (2 ( 2 cos 2, g (3 ( 4 sin 2, g (4 ( 8 cos 2, g (5 ( 6 sin 2, g (6 ( 32 cos 2, g (7 ( 64 sin 2, g (8 ( 28 cos 2. Note Remembering that 2 sin cos sin 2 makes life a little easier. Otherwise, g ( ( 2 cos sin g (2 ( 2 sin 2 2 cos 2, (Prouct Rule g (3 ( 4 sin cos + 4 cos sin 8 sin cos 4g ( (. Thus further erivatives can be epresse in terms of either g ( of g (2. 2
3 5 (2006 Differentiate when πl/2 for any l Z. : ( sin + cos 6 (2007 Differentiate sin + cos, ( (sin + (cos ( (sin + (sin + (cos i ln 2 ( +, for >, (sin 2 sin sin cos sin 2 + sin cos 2 sin (cos 2 sin cos2 cos 3 + sin 3 sin 2 cos 2 4 sin cos3 sin 2. 2 by the Chain Rule cos This can be one in two ways. First the Prouct Rule gives ln2 ( + (ln ( + ln ( + ln ( + ln ( + + ln ( + + ln ( + + ln ( ln ( +. + ln ( + 3
4 ii Alternatively, the Chain Rule gives for >, 5 times. Let ln2 ( + y y2, where y ln ( +, y ln ( + 2y 2 ln ( + ln f ( ln ( +, + ( +. 2 ln ( +. + There are two ways you coul look at the first erivative. First you coul use the Chain Rule: ( + ln ln y y y, where so Thus y + ( + ln ( + 2 y 2 ( 2. ln y y y y + 2, 2 ( 2 2 ( ( +. ( Alternatively, you can use the properties of logarithm to write ( + f ( ln ln ( + ln (. 4
5 Then the Sum Rule gives ( + ln ln ( + ln ( + +. (2 Check that answers ( an (2 are the same. To continue, it is simpler to start from (2 rather than (: f ( ( f (3 ( f (4 ( f (5 ( + +, so f (2 ( 2 ( ( + 3, 6 ( ( + 4, 24 ( ( + 5. ( 2 + ( + 2, iii tan 2, when (2k + π/2 for any k Z. : It is most straightforwar to use the Prouct Rule but as an eample of the Quotient Rule: tan 2 ( sin 2 cos 2 cos2 sin2 sin 2 cos2 (cos 2 2 cos2 (2 sin cos sin 2 ( 2 cos sin cos 4 2 sin cos2 + sin 3 cos 3 2 sin ( cos 2 + sin 2 cos 3 2 tan cos 2. 5
6 having use sin 2 + cos 2. Further use of trigonometric ientities allow the answer to be written as 2 (tan ( + tan 2. iv e (sin + cos, 6 times, Let f ( e (cos + sin. Then f ( ( e (cos + sin + e ( sin + cos 2e cos, f (2 ( 2e (cos sin, f (3 ( 2e (cos sin + 2e ( sin cos 4e sin, f (4 ( 4e sin 4e cos 4f (, f (5 ( 4f ( ( 8e cos, f (6 ( 4f (2 ( 8e (cos sin. Notice that once we have a relation such as f (4 ( 4f (, then all subsequent erivatives are just multiples of erivatives alreay foun. 7 (2008 Differentiate i 2 2, 3 times. Recall that 2 e ln 2. Thus 2 e ln 2 ey y Write f ( 2 2, then y, y ( ln 2 e ey, where y ln 2, (Chain Rule, e ln 2 ln 2 2 ln 2. f ( ( 2 ln 2 2, f ( ( 2 ln 2 2 2, f (3 ( 2 ln
7 ii for, 5 times. e +, Alternatively, let f ( e / ( + so ( + f ( e an ifferentiate this 5 times. Write f ( e / ( + to get the following: f ( ( f (2 ( f (3 ( f (4 ( f (5 ( e + e ( + 2 e ( + 2, e + 2 e ( e ( + 3 ( + 2 e ( + 3, e + 3 e ( e ( e e + 4 e ( e e ( ( + 5, e + 5 e ( e e 20 ( + 6 e ( ( + 6. ( + 4 ( e ( + 4, ( e ( e ( + 5 ( e ( e The suggeste alternative is: write ( + f ( e. repeate ifferentiation gives ( + f ( ( + f ( e, ( + f (2 ( + 2f ( ( e, ( + f (3 ( + 3f (2 ( e, ( + f (4 ( + 4f (3 ( e, ( + f (5 ( + 5f (4 ( e. ( + 5 In which case These coul be solve to give the erivatives. 7
8 iii sinh + cosh, 6 times : Let h ( sinh + cosh. Then h ( ( 2 sinh + cosh, h (2 ( 3 cosh + sinh, h (3 ( 4 sinh + cosh, h (4 ( 5 cosh + sinh, h (5 ( 6 sinh + cosh, h (6 ( 7 cosh + sinh. 8 (2009 Differentiate i 2 sin π 2. for 0. ( 2 sin π 2 ( 2 sin π + (sin π by the Prouct Rule 2 sin π sin y y y, where y π, 2 by Chain Rule, 2 sin π ( cos y 2π 3 2 sin π 2 2π cos π 2. 8
9 ii ln 2 ( +, for >, 5 times, In Question 6 i you have alreay been aske to ifferentiate this function once. The further erivatives are: f ( ln 2 ( +, f ( ln ( + ( 2 +, f (2 ( 2 ( + 2 2ln ( + ( + 2, f (3 6 ( + ( 3 + 4ln ( + ( + 3, f (4 ( 22 ( + 4 2ln ( + ( + 4, f (5 ( 00 ( ln ( + ( + 5. Practice of ifferentiation. On the basis that you learn mathematics only by oing mathematics here are a series of questions on ifferentiation to help you learn, or perhaps refresh, your facility on ifferentiating stanar functions. 9 Use the Prouct Rule to ifferentiate i e 2 (e 2, e2 (e 2 (e e e e + e e e e + e e 2e 2. (Prouct Rule 9
10 ii e ln, for > 0, e ln e iii (sin (ln, for > 0, ln + e ln e ln + e e ln +. sin ((sin (ln ln + sin ln (Prouct Rule (Prouct Rule cos ln + sin. iv e cos ln, for > 0, (e cos ln (e (cos ln e (cos ln + e (cos ln e (cos ln + e ( cos ln ln + cos (Prouct Rule by a secon application of the prouct rule, e cos ln e sin ln + e cos. 0
11 0 Use the Quotient Rule to ifferentiate i e /e, e ( e e e (e 2 (Quotient Rule e (e 2 e e. ii e /, for 0, iii /e, ( e ( e iv ln / 2, for > 0, ( ln 2 e e 2 e e 2 2 e. (Quotient Rule e e (e 2 (Quotient Rule e e (e 2 e. 2 ln 2 2 ln ( 2 2 (Quotient Rule ln (2 ( ln 3.
12 v 2 / ln, for >, ( 2 ln ln 2 2 ln (ln 2 (Quotient Rule 2 ln 2 ln 2 vi / tan 2, when πl for any l Z. Using the Quotient Rule we have ( tan 2 (2 ln ln 2. ( cos 2 sin 2 sin2 cos2 cos 2 sin2 ( sin sin2 cos sin 2 cos 2 sin cos sin 4 ( sin 2 + cos 2 2 cos 2 cos sin 3 sin 3 Using instea the Chain Rule let u tan 2 which we have ifferentiate in Question 6iii as u 2 tan cos 2. Then ( tan 2 ( u u u 2 2 tan cos 2 which leas to the same result as before. tan 2 tan 4 cos 2, ( u u 2
13 vii (sin + cos / (sin cos, when ( + 4k π/4 for any k Z. ( sin + cos sin cos (sin +cos (sin cos (sin + cos (sin cos 2 (sin cos (cos sin (sin cos 2 by the Quotient Rule (sin + cos (cos + sin (sin cos 2 sin cos sin2 cos 2 + cos sin (sin cos 2 (sin cos sin cos + sin2 + cos 2 + cos sin (sin cos 2 2 sin 2 2 sin cos + cos 2 2 sin 2. Use the Chain Rule to ifferentiate i e 2, Let u 2 so e 2 e u. Then ii e, e 2 eu eu u u e u (2 2e. Let u so e e u. Then (Chain Rule, e eu eu u u e u ( (Chain Rule, e. 3
14 iii cos (/, for 0, Let u / so cos (/ cos u. Then cos (/ cos u cos u u u sin u ( ( sin ( 2 (Chain Rule, sin (/ 2. iv sin (ln, for > 0, Let u ln so sin (ln sin u. Then sin (ln v ln /2, for >, sin u cos u ln sin u u u cos (ln. Let u ln so ln /2 u /2. Then by the Chain Rule, ln /2 u/2 vi ln ( + cos ( /2, for > 0, u/2 u u 2u /2 ln 2 ln /2. by the Chain Rule, This is a function of a function of a function of. Thus given we first calculate v /2. Net we calculate u + cos v ( + cos ( /2 4
15 an finally the function we wish to ifferentiate is ln u. So ln ( + cos ( /2 ln u ln u u u ( + cos ( /2 u u u ( + cos v v v (Chain Rule, ( + cos v u by the Chain Rule, again, /2 ( sin v ( ( sin v u 2 /2 sin ( /2 2 /2 ( + cos ( /2, on substituting for u an v. +sin vii e 2, Let w sin, v + w 2, u +sin v so e 2 e u. Then by the Chain Rule applie three times, e +sin eu eu u v w u v w e u v v e u 2 2w cos v ( + w 2 sin w sin cos + sin 2 e +sin 2 2 Differentiate sin sin 2 e +sin 2. i sin ( ln, for > 0. Let y ln when, by the Prouct Rule y ln ln ln + 5 ln +.
16 Then, by the Chain Rule, sin ( ln sin y y y cos y (ln + (ln + cos ( ln. ii ( /2 sin 2 / ( 2 +, for > 0, ( /2 sin (2 + /2 sin2 /2 sin 2 (2 + ( by Quotient Rule ( ( 2 + sin 2 /2 + /2 sin2 /2 (2 sin 2 ( by Prouct Rule ( ( 2 sin /2 sin cos 2 3/2 sin 2 2 /2 ( (2 + ( sin sin cos 4 2 sin 2 2 /2 ( iii (sin / 2 for 0, Let y sin / when, by the Quotient Rule, y ( sin sin sin 2 cos sin 2. So, by the Chain Rule (sin / 2 y2 y y 2y ( cos sin 2 2 sin ( cos sin 3. 6
17 iv sin (, for 0, The Prouct Rule gives ( sin ( sin ( + ( sin (. For the secon term let y / an use the Chain Rule ( ( sin sin y y y cos y ( ( cos. 2 Combine to get ( ( ( sin sin ( v m sin (, m, n an 0, n cos ( ( m m sin sin n n + m sin ( sin n vi sin m cos n, m, n, By the Prouct Rule. ( n m m + m sin y y y where y / n ( ( sin m m + m cos y n ( m m sin n n n+ ( + n m n cos n (sinm cos n cos n sinm + sin m cosn. m cos n+ sin m n sin m+ cos n sin m cos n ( m cos 2 n sin 2. 7
18 vii 2 sin cos 2, Let y 2 sin cos 2 so y ( 2 sin cos 2 4 sin cos 8 cos sin 4 sin cos 2 sin 2. Then, by the Chain Rule 2 sin cos 2 y y viii m ln n, m, n an > 0, By the Prouct Rule, y 2 ( 2 sin 2 y sin 2 2 sin cos 2. (m ln n ln n m + m lnn. For the secon term use the Chain Rule, so let y ln, then ln n Combining we have i e ln for >, yn y y nyn ln n lnn. (m ln n m m ln n + m n lnn m ln n (m ln + n. Let v ln an u v ln. Then the Chain Rule, applie twice, gives e ln eu u v u v v ln eu v e u 2 v e ln 2 ln. 8
19 e / ln for >, Let v ln an u v /2 / ln. Then the Chain Rule, applie twice, gives i ln ( + + e. ln e u u v e/ u v eu v /2 v e u 2v 3/2 ln e/ 2 ln 3/2. ln Let v + e an u + v + + e. Then the Chain Rule applie twice gives ( ln + + e ln u u u v v u u 2 v e e 2 + e ( + + e e 2 ( + e + + e ( + v ( + e v 9
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