Ребятки помогите пожалуйста с подробным решением. Created by aykam98. matematika-ru

Ребятки помогите пожалуйста с подробным решением...

aykam98 @aykam98

November 2021 1 2 Report

Ребятки помогите пожалуйста с подробным решением

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logophobia

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Производные:

1) y' = (5x^6)' = 5*6x^(6–1) = 30x^5 ;

2) y' = (2xsinx)' = 2sinx + 2xcosx ;

3) y' = (4cosx/x^5)' = ( – 4 x^5 sinx – 5 x^(5–1) cosx ) / x^(2*5) =
= – ( 4 x sinx + 5 cosx ) / x^6 ;

4) y' = √[ 6x + 2 ] ' = ( ( 6x + 2 )^(1/2) )' = (1/2)( ( 6x + 2 )^(1/2–1) ) * 6 =
= 3 / ( 6x + 2 )^(1/2) = 3 / √[ 6x + 2 ] ;

5) y' = ( ( 7x + 3 )^3 )' = 3 ( ( 7x + 3 )^(3–1) ) * 7 = 21(7x+3)² ;

6) y' = (1/x)' = ( x^(–1) )' = (–1) ( x^(–1–1) ) = – x^(–2) = –1/x² ;

7) y' = (1/x³)' = ( x^(–3) )' = (–3) ( x^(–3–1) ) = –3x^(–4) = –3/x^4 ;

Интегралы:

1) ∫ 2x√x dx = ∫ 2x x^(1/2) dx = 2 ∫ x^(3/2) dx = 2/(3/2+1) x^(3/2+1) + C =
= 2/(5/2) x^(2+1/2) + C = (4/5) x²√x + C ;

2) ∫ (1/x+1/x³) dx = ∫ dx/x + ∫ x^–3 dx = ln|x| + x^(–3+1)/(–3+1) + C =
= ln|x| –(x^–2)/2 + C = ln|x| – 1/(2x²) + C ;

3) ∫ (sinx+x³) dx = ∫ sinx dx + ∫ x³dx = –cosx + (x^(3+1))/(3+1) + C =
= (x^4)/4 – cosx + C ;

4) ∫ 4cosx dx = 4 sinx + C ;

5) ∫ (x²+3)²dx = ∫ ( x^4 + 6x^2 + 9 ) dx = ∫ x^4 dx + 6 ∫ x²dx + 9 ∫ dx =
= (x^(4+1))/(4+1) + 6 (x^(2+1))/(2+1) + 9x + C = (x^5)/5 + 6 (x^3)/3 + 9x + C ;

3 votes Thanks 1

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aykam98 July 2022 | 0 Ответы

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