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Реферат: Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.15

Задача 15 . Найти производную .

15.1.

x'= 6t*t3 -3t2 (3t2 +1) = -t2 -1

3t6 t4

y'= cos(t3 /3+t)(t2 +1)

y'x = cos(t3 /3+t)(t2 +1)t4 = -t4 cos(t3 /3+t)

-t2 -1

15.2.

x'= -t _

√(1-t2)

y'= 1 _

2√(1+t)cos2 √(1+t)

y'x = -√(1-t2 ) = -√(1-t2 ) _

2t√(1+t)cos2 √(1+t) 2tcos2 √(1+t)

15.3.

x'= 1-t _

√(2t-t2 )

y'= 2 _

3 3 √(1-t)5

y'x = 2√(2t-t2 ) = 2√(2t-t2 ) _

3 3 √(1-t)5 (1-t) 3 3 √(1-t)2 (1-t)2

15.4.

x'= cost = 1

√(1-sin2 t)

y'= sint = 1

√(1-cos2 t)

y'x = 1

15.5.

x'= 1+t/√(t2 +1) = 1 _

t+√(t2 +1) √(t2 +1)

y'= √(t2 +1)+ t2 = 2t2 +1_

√(t2 +1) √(t2 +1)

y'x = (2t2 +1)√(t2 +1) = 2t2 +1

√(t2 +1)

15.6.

x'= 1-t _

√(2t-t2 )

y'= 1 = 1 _

√(1-(t-1)2 ) √(2t-t2 )

y'x = √(2t-t2 ) = 1_

√(t2-t2 )(1-t) 1-t

15.7.

x'= -2et _ = -2et _

sin2 (2et ) 4sin2 et cos2 et

y'= et = et _

tget cos2 et sinet coset

y'x = 4et sin2 et cos2 et = -2sinet coset

-2et sinet coset

15.8.

x'= -1 = -1 _

ctgt sin2 t sint cost

y'= 2sint

cos3 t

y'x = -cos3 t = -1/2*ctg2 t

2sin2 tcost

15.9.

x'= et/2 _

2(1+et )

y'= et _

2√(1+et )

y'x = 2et (1+et ) = √(et +e2t )

2et/2 √(et +1)

15.10.

x'= √(1+t) *√(1+t) *-1-t-1+t = -1_

√(1-t) 2√(1-t) (1+t)2 1-t2

y'= -t _

√(1-t2 )

y'x = t(1-t2 ) = t√(1-t2 )

√(1-t2 )

15.11.

x'= 2t3_

1-t4

y'= (1+t2 )(-2t(1+t2 )-2t(1-t2 )) = -2

√(1+2t2 +t4 -1+2t2 -t4 )(1+t2 )

y'x = -2(1-t4 ) = t4 -1

2t3 t3

15.12.

x'= -t _

√(1-t2 )

y'= √(1-t2 )+t2 /√(1-t2 ) = 1 _

1-t2 (1-t2 )3/2

y'x = -√(1-t2 )_ = 1_

t(1-t2 )3/2 t3 -t

15.13.

x'= -t = -1 _

√(1-1+t2 )√(1-t2 ) √(1-t2 )

y'= -2arccost

√(1-t2 )

y'x = 2arccost√(1-t2 ) = 2arccost

√(1-t2 )

15.14.

x'= √(1-t2 )+t2 /√(1-t2 ) = 1 _

1-t2 (1-t2 )3/2

y'= t *-t2 /√(1-t2 )-1-√(1-t2 ) = -1

1+√(1-t2 ) t2 t

y'x = √(1-t2 )

t2

15.15.

x'= -4(1+cos2 t)costsint

y'= -sin3 t-2cos2 tsint = -1-cos2 t

sin4 t sin3 t

y'x = 1+cos2 t = 1 _

4sin3 t(1+cos2 t)costsint 4sin4 tcost

15.16.

x'= (1+t)(-1-t-1+t) = -2_

(1-t)(1+t)2 1-t2

y'= -t _

√(1-t2 )

y'x = -t (1-t2 )_ = t√(1-t2 )

-2√(1-t2 ) 2

15.17.

x'= 1 = 1 _

t2 √(1-1/t2 ) t√(t2 -1)

y'= t + t = 2t _

√(t2 -1) √(t2 -1) √(t2 -1)

y'x = 2t2 √(t2 -1) = 2t2

√(t2 -1)

15.18.

x'= 1_

tln2 t

y'= t *-t2 /√(1-t2 )-1-√(1-t2 ) = -1

1+√(1-t2 ) t2 t

y'x = -tln2 t = -ln2 t

t

15.19.

x'= 1 _

2√t√(1-t)

y'= 1 _

4√t√(1+√t)

y'x = 2√t√(1-t) = √(1-√t)

4√t√(1+√t) 2

15.20.

x'= 2arcsint

√(1-t2 )

y'= √(1-t2 )+t2 /√(1-t2 ) = 1 _

1-t2 (1-t2 )3/2

y'x = √(1-t2 ) = 1 _

2(1-t2 )3/2 arcsint 2(1-t2 )arcsint

15.21.

x'= √(t2 +1)+t2 /√(t2 +1) = 2t2 +1

√(t2 +1)

y'= t *-t2 /√(1-t2 )-1-√(1-t2 ) = -1

1+√(1-t2 ) t2 t

y'x = -√(t2 +1) = -1 _

(2t2 +1)√(t2 +1) 2t2 +1

15.22.

x'= 1/(1+t2 )

y'= (t+1)(t(t+1)/√(t2 +1)-√(1+t2 )) = t-1 _

√(1+t2 )(1+t) √(t2 +1)(1+t)

y'x = (t-1)(1+t2 ) = (t-1)√(1+t2 )

√(1+t2 )(t+1) 1+t

15.23.

x'= -2t/(1-t2 )

y'= -t = -1/√(1-t2 )

√(1-1+t2 )√(1-t2 )

y'x = 1-t2 = √(1-t2 )

2t√(1-t2 ) 2t

15.24.

x'= (t-1)2 (t-1-t-1) = -1_

((t-1)2 +(t+1)2 )(t-1)2 t2 +1

y'= -t = -1/√(1-t2 )

√(1-1+t2 )√(1-t2 )

y'x = t2 +1_

√(1-t2 )

15.25.

x'= √(1+sint)√(1+sint)(-cost(1+sint)-cost(1-sint)) = -1_

2√(1-sint)√(1-sint)(1+sint)2 cost

y'= tgt/cos2 t-tgt= tg3 t

y'x = -tg3 tcost

15.26.

x'= 1-2t _ t√t(-t-1+t) = √(1-t)

2√(t-t2 ) 2(t+1-t)√(1-t)t2 2√t

y'= 1 + arcsin√t _ √(1-t) = arcsin√t

2√t 2√(1-t) 2√t√(1-t) 2√(1-t)

y'x = 2√t arcsin√t = √t arcsin√t

2(1-t) (1-t)

15.27.

x'= 1 = 1 _

tgtcos2 t sintcost

y'= -2cost

sin3 t

y'x = -2cost = -2_

sin4 tcost sin4 t

15.28.

x'= (2tlnt+t)(1-t2 )+2t3 lnt 2t = 2tlnt

(1-t2 )2 2√(1-t2 ) (1-t2 )

y'= √(1-t2 )+t2 /√(1-t2 ) arcsint + t/(1-t2 ) – t/(1-t2 ) = arcsint

1-t2 (1-t2 )3/2

y'x = arcsint(1-t2 )2 = arcsint√(1-t2 )

2tlnt(1-t2 )3/2 2tlnt

15.29.

x'= 2esec^2t sec2 t tgt= 2esec^2t sint

cos3 t

y'= lncost _ sint + 1/cos2 t-1= lncost-sintcost+sin2 t

cos2 t cost cos2 t

y'x = 1/2*e-sec^2t ctgt(lncost-sintcost+sin2 t)

15.30.

x'= √(1-t2 )+t2 /√(1-t2 ) arcsint + t/(1-t2 ) – t/(1-t2 ) = arcsint

1-t2 (1-t2 )3/2

y'= √(1-t2 )+t2 /√(1-t2 ) = 1 _

1-t2 (1-t2 )3/2

y'x = (1-t2 )3/2 = 1 _

(1-t2 )3/2 arcsint arcsint

15.31.

x'= 1+t/√(t2 +1) = 1 _

t+√(t2 +1) √(t2 +1)

y'= t _ t *-t2 /√(1-t2 )-1-√(1-t2 ) = t + 1 = t2 +2√(1-t2 )

2√(1+t2 ) 1+√(1-t2 ) t2 2√(1+t2 ) t 2t√(1-t2 )

y'x = (t2 +2√(1-t2 ))√(1+t2 )

2t√(1-t2 )