Решения к Сборнику заданий по высшей математике Кузнецова Л. А. – 2. Дифференцирование. Зад.6


Задача 6 . Найти производную.

6.1.

ex + 2 e 2 x + e x

Y’ = 1- √( e 2 x + e x +1) = 2+ e x +√( e 2 x + e x +1)- e x √( e 2 x + e x +1)-2 e 2 x – e x =

2+ex +2√(e2x +ex +1) 2+ex +2√(e2x +ex +1)

= (2-e x )√(e 2x +e x +1)+2+e x -2e x

2+ex +2√(e2x +ex +1)

6.2.

y’ = 1/4*e2x (2-sin2x-cos2x)+1/8*e2x (-2cos2x+2sin2x)=1/8*e2x (4-2sin2x-2cos2x-2cos2x+2sin2x)=1/8*e2x (4-4cos2x)=e2x *sin2x

6.3.

y’ = 1 * 1 * 2e x = e x.

2 1 + (e x -3) 2 4 e2x -6ex +10

4

6.4.

y’ = 1 * 1-2 x * -2 x ln2(1+2 x )-(1-2 x )2 x ln2 = (2 x -1)2 x ln4 = 2 x (2 x -1)

ln4 1+2x (1+2x )2 ln4(1+2x )3 (1+2x )3

6.5.

e x (√(e x +1)+1) _ e x (√(e x +1)-1)

y’ = e x + √(e x +1)+1 * 2√(e x +1) 2√(e x +1) =

√(ex +1) √(ex +1)-1 (√(ex +1)+1)2

= e x + e x √(e x +1)+e x – e x √(e x +1)+e x = √(ex +1)

√(ex +1) 2ex √(ex +1)

6.6.

Y’ = 2/3*3/2*√(arctgex ) * e x = e x √(arctge x )

1+ex 1+ex

6.7.

y’ = 2e x – 2e x = e x

2(e2x +1) 1+e2x 1+e2x

6.8.

6.9.

Y’ = 2/ln2*((2x ln2)/(2√(2x -1))-(2x ln2)/(1+2x -1))=2x/√(2x -1)-2

6.10.

e x (√(1+ e x )+1) _ e x (√(1+ e x )-1)

Y’= 2√(1+ex )+2 e x ( x -2) _ √(1+ e x )+1 * 2√(1+ e x ) 2√(1+ e x ) =

2√(1+ex ) √(1+ex )-1 (√(1+ex )+1)

= xe x +2 _ 2e x √(1+e x )+2e x = xe x.

√(1+ex ) ex √(1+ex )( √(1+ex )+1) √(1+ex )

6.11.

Y’= αe αx (αsinβx-βcosβx)+e αx (αβcosβx+β 2 sinβx) =

α2 +β2

= e αx (α 2 sinβx+β 2 sinβx) = eαx sinβx

α2 +β2

6.12.

Y’= αe αx (βsinβx-αcosβx)+e αx (β 2 cosβx+αβsinβx) =

α2 +β2

= e αx (β 2 cosβx+2αβsinβx-α 2 cosβx)

α2 +β2

6.13.

Y’= aeax * ┌ 1 + acos2bx+2bsin2bx ┐+ eax ┌ -2absin2bx+4b 2 cos2bx ┐=

└ 2a 2(a2 +4b2 ) ┘ └ 2(a2 +4b2 ) ┘

= eax /2*(1+cos2bx)= eax cos2 bx

6.14.

y’ = 1 – e x – e x = 1 – e x – e x – e 2x = 1 + e 2x.

(1+ex )2 1+ex (1+ex )2 (1+ex )2

6.15.

3/6*ex/6 *√(1+ex/3 ) + 1/3*e x/3 (1+e x/6 )

Y’= 1 – 2√(1+e x/3 ) _ 3/6*e x/6 =

(1+ex/6 )√(1+ex/3 ) 1+ex/3

= 1- e x/6 +e x/2 +e x/3 +e x/2 _ e x/6 = 1- e x/3 – e x/6 .

2(1+ex/6 )(1+ex/3 ) 2(1+ex/3 ) 2(1+ex/6 )(1+ex/3 )

6.16.

y’ = 1 – 8e x/4 = 1 – 2e x/4 .

4(1+ex/4 )2 (1+ex/4 )2

6.17.

ex + e 2x

Y’= √(e 2x -1) ­_ e – x = e x (e x +√(e 2x -1)) _ e – x *e x = e x -1 .

ex +√(e2x -1) √(1-e-2x ) (ex +√(e2x -1))√(e2x -1) √(e2x -1) √(e2x -1)

6.18.

e 2x

Y’= 1+e-x arcsinex – e – x *e x + √(1-e 2x ) =

√(1-e2x ) 1+√(1-e2x )

= 1+e-x arcsinex – 1 + e 2x =

√(1-e2x ) (1+√(1-e2x )) √(1-e2x )

= e-x arcsinex

6.19.

Y’= 1- e x +e-x/2 arctgex/2 – e – x/2 *e x/2 _ ex/2 arctgex/2 =

1+ex 1+ex 1+ex

= 1- ex + 1 + arctgex/2 * 1-ex = arctgex/2 * 1-ex.

1+ex 1+ex ex/2 (1+ex ) ex/2 (1+ex )

6.20.

Y’= 3×2 ex3 (1+x3 )-3ex3 x2 = 3×5 ex3

(1+x3 )2 (1+x3 )2

6.21.

Y’= b *memx √a = emx.

m√(ab)(b+ae2mx ) √b b+ae2mx

6.22.

Y’= e3^√x /3√x(3 √x2 -23 √x+2)+3e3^√x (2/(33 √x)-2/(33 √x2 ))= e3^√x

3^√x= кубический корень из х

6.23.

( ex +2e2x _ ex )(√(1+ex +e2x )-ex +1) _ ( ex +2e2x _ ex )(√(1+ex +e2x )-ex -1)

Y’= √(1+ex +e2x )-ex +1 * 2√(1+ex +e2x ) 2√(1+ex +e2x ) =

√(1+ex +e2x )-ex -1 (√(1+ex +e2x )-ex +1)2

= ex (1+2e2x -2√(1+ex +e2x )) = 1 .

(ex (1+2e2x -2√(1+ex +e2x )))√(1+ex +e2x ) √(1+ex +e2x )

6.24.

Y’= cosxesinx (x-1/cosx)+esinx (1-sinx/cos2 x)= esinx (xcosx-sinx/cos2 x)

6.25.

Y’= ex /2((x2 -1)cosx+(x-1)2 sinx)+ex /2(2xcosx-(x2 -1)sinx+2(x-1)sinx+(x-1)2 cosx)=

= ex /2(x-1)(5x+3)cosx

6.26.

Y’= ex +e-x = e3x +ex.

1+(ex – e-x )2 e4x – e2x +1

6.27.

Y’= e3^√x /3 √x2 (3 √x5 -53 √x4 +20x-603 √x2 +1203 √x-120)+e3^√x (53 √x2 -203 √x+20-120/3 √x+120/3 √x2 )= e3^√x (x-40)

6.28.

Y’= -3e3x sh3 x+3e3x sh2 xchx = e3x (chx-shx)

3sh6 x sh4 x

6.29.

Y’= – e-x + e2x = √(e4x – e2x )-√(e-2x -1) = √(e2x -1)-√(1-e2x )

√(1-e-2x ) √(1-e2x ) √(1-e-2x )*√(1-e2x ) √(1-e-2x ) √(1-e2x )

6.30.

Y’= xe-x2 (x4 +2×2 +2)-1/2*e-x2 (4×3 +4x)= x5 e-x2

6.31.

Y’= 2xex2 (1+x2 )-2ex2 x = 2×3 ex2

(1+x2 )2 (1+x2 )2



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