Differentiation 9F. tan 3x. Using the result. The first term is a product with. 3sec 3. 2 x and sec x. Using the product rule for the first term: then
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1 Differentiation 9F a tan Using the reslt tan k k sec k sec 4tan Let tan ; then 4 sec and sec tan sec d tan tan The first term is a proct with and v tan and sec Using the proct rle for the first term: sec sec tan sec sec tan c tan( ) sec ( ) a cot 4 Let 4; then cot 4 and cosec cosec 4 4 cosec 4 sec5 Let 5; then sec 5 and sectan 5sec tan 5sec5 tan 5 Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free.
2 c cosec 4 Let 4; then cosec 4 and cosec cot f sec This is a qotient, so let sec and v and se the qotient rle. sec (sec tan ) and d 4coseccot 4cosec 4cot 4 sec (sec ) Let sec ; then g d (sec tan ) sec sec ( tan ) cosec sec tan and Let cosec ; then cosec cot and sec tan secsec tan 6sec tan e cot This is a proct, so let and v cot and se the proct rle. and cosec cot cosec ( cosec ) cot h ( cosec cot ) 6cosec cosec cot 6cosec cot cot ( ) Let cot( ); then cosec ( ) and cosec () 4cot( ) cosec ( ) a f ( ) (sec ) f ( ) (sec ) sec tan (sec ) tan Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free.
3 c d f ( ) cot (cot ) f ( ) (cot ) ( cosec ) (cot ) cosec f ( ) cosec (cosec ) f ( ) (cosec ) ( cosec cot ) cosec cot f ( ) tan (tan ) f ( ) tan sec tan sec c f ( ) tan Let and v tan and sec Using the qotient rle, tan sec f ( ) tan d f ( ) e sec Let e and v sec e and sec tan Using the proct rle, f ( ) e sec tan e sec e sec (tan ) e f 4 a f ( ) sec (sec ) f ( ) (sec ) sec tan sec tan f ( ) cot (cot ) f ( ) (cot ) ( cosec ) f ( ) cot sec cosec Let and v sec and sec tan Using the proct rle, f ( ) sec tan sec tan f ( ) Let tan and v sec and Using the qotient rle, sec tan f ( ) e f ln f ( ) tan Let ln and v tan and sec Using the qotient rle, tan ln sec f ( ) tan tan ln sec tan tan e f ( ) cos tan Let e and v cos tan e sec and sin Using the qotient rle, tan tan e sec cos e ( sin ) f ( ) cos tan tan e sec e sin cos tan e (sec sin ) cos tan e (sec sec tan ) tan e sec (sec tan ) Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free.
4 5 a sec cosec cos sin Let sec and v cosec sec tan and cosec cot Using the proct rle, sec ( cosec cot ) cosec (sec tan ) cos sin cos sin sin sin cos cos sin cos c π When, 4 4 π π cos sin or, sing the alternative epression, 8 4 d Alternative soltion: = cosec cos sin sin (ecase sin sin cos ) 4cosec cot At stationar points d 0 0 cos sin cos sin tan tan In the interval 0 < π there are two soltions, π and π. 4 4 So the nmer of stationar points is. Alternative soltion: 4cosec cot 0 cosec 0 t cot 0 has two soltions in the interval 0 < π. So there are stationar points. 6 7 Eqation of tangent is 4 8 π 8π π 0 This is in the reqired form a c 0 With a = 4, = 9 and c 8π. sec cos Let and v cos 0 and sin Using the qotient rle, cos 0 ( sin ) cos sin sec tan cos cot tan Let and v tan 0 and sec Using the qotient rle, tan 0 sec sec tan tan cos cosec sin sin cos Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free. 4
5 8 a Let arccos So cos sin sin sin cos sin cos Let arctan So tan sec sec tan 9 a Let arccos Let t ; then arccos t dt and dt t dt dt t 4 Let arctan Let t ; then arctan t dt and dt t dt dt t 4 4 c Let arcsin So sin sin cos cos sin cos cos sin ( ) 9 9 d Let arccot So cot cosec cosec cot Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free. 5
6 9 e Let arcsec So sec sec tan sec tan tan sec tan sec f Let arccosec So cosec cosec cot cosec cot cot cosec cot cosec g Let arcsin Let then arcsin, Using the qotient rle, d ( ) d ( ) ( ) ( ) ( ) ( ) ( ) d ( ) ( ) ( ) ( ) h Let arccos Let t and arccos t dt and dt t dt dt t 4 i Let e arccos Using the proct rle, e e arccos e arccos j Let arcsin cos Using the proct rle, arcsin ( sin ) cos cos sin arcsin k Let arccos Using the proct rle, arccos arccos arccos Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free. 6
7 9 l Let 0 a e arctan Let arctan ; then e e and e e arctan arctan Let arctan and v and 4 Using the qotient rle, arctan 4 arctan ( 4 ) When, arctan 4 π 4 4 4π 4π 9 9 When, π arctan π π 9 Eqation of normal is π 9 9 4π 9 9 π 4π 6 8π 9 (arccos ) arccos cos( ) Let ; then cos and sin sin sin sin cos sin cos cos a cosec5 5cosec5 cot 5 5cosec5 cot 5 cot 5 cosec 5 cot 5 cosec 5 5 Pearson Ecation Ltd 07. Coping permitted for prchasing instittion onl. This material is not copright free. 7
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