Задача 9. Найти производную.
9.1. />
y’= 2(1/cos2x+1/sin2x)= 2cos2xsin2x = 2 _
2+(tgx-ctgx)2(sin4x+cos4x)sin2xcos2x sin4x+cos4x
9.2. />
√(5x)_ √5(√x-2)
y’= 1 * 2√x 2√x = √5 = √5 _
√(1-(√x-2)2/(5x)) 5x 5x√(5x-(√x-2)2) 10x√(x+√x-1)
9.3. />
y’= 1/2*√(2+x-x2) – (2x-1)(1-2x)+ 54 = 9 _
8√(2+x-x2) 24√(9-(2x-1)2) 4√(2+x-x2)
9.4. />
x2— √(1+x2)+1
y’= x2* √(1+x2) = √(1+x2) – 1 _
x2+(√(1+x2)-1)2×22√(1+x2)(x2-√(1+x2)+1)
9.5. />
2x√(x4+16) – 2x(x2-4)
y’= — 1 * √(x4+16)= -√(x4+16) *
√(1-(x2-4)2/(x4+16)) x4+16 √(x4+16-x4+8×2-16)
* 2x(x4+16)-2x(x2-4)= x2-x4-20 _
(x4+16)√(x4+16) √2(x4+16)
9.6. />
3√(6x)-3(3x-1)
y’= √(2/3) * √(6x) = 6x√2(18x-9x+3) = 3x+1 _
1+(3x-1)2/(6x) 6x 6x√3√(6x)√(6x+9×2-6x+1) √x(9×2+1)
9.7. />
y’= x+1 * x+1-x+1_ 1 = 1_
4(x-1) (x+1)22+2x2x4-1
9.8. />
y’= 1/2*√(8x-x2-7)+(x-4)(4-x) +9√6 = 8x-23 +9 _
2√(8x-x2-7) 2√(6x)√(6-x-1) 2√(8x-x2-7) 2√x√(5-x)
9.9. />
x2arctg√x+(1+x)x2_— 2x(1+x)arctg√x
y’= 2√x(1+x) _ √x+x/(2√x)_= –PAGE_BREAK–
x43x3
= 1 _ 2arctg√x_ 2arctg√x_ 1 _
2×2√x x3x23x2√x
9.10. />
y’= x2arccosx — x3_ 2x√(1-x2)+ x(2+x2)=
3√(1-x2) 9 9√(1-x2)
= x2arccosx – 3×3+2x(1-x2)-2x-x3= x2arccosx
9√(1-x2)
9.11. />
y’= — 1 + (x-1-x)arctg√x+ x+1= — arctg√x
4√x32x24x√x 2×2
9.12. />
y’= 1/2*√(x(2-x))+(3+x)(2-2x)_ 3_=
4√(x(2-x)) 4√(x/2)√(1-x/2)
= 2x(2-x)+2(3+x)(1-x)_ 3 _= -x2_
4√(x(2-x)) 2√(x(2-x)) √(x(2-x))
9.13. />
y’= (4×3*x3-3×2(4+x4))arctg(x2/2)+ (4+x4)x– 4/x2=
x6x3
= (x4-12)arctg(x2/2)+ 4(x4-3)
x4x2(4+x4)
9.14. />
y’= √x(x+1-x) +1 _= -1 _
2√(1-x/(x+1))√(x+1)(x+1)22(x+1)√x 2(x+1)2√x
9.15. />
y’= -2 + 2x2_ 2arccosx= -x+2x _ arccosx= 1 _ arccosx
4×3√(1/x2-1) 2×4√(1-x2) 2x72x3√(1-x2) x72x2√(1-x2) x7
9.16. />
y’= 6 – x/2*√(x(4-x)) – (6+x)(4-2x)= 3 – x/2*√(x(4-x)) – (6+x)(2-x)=
4√x√(1-x/4) 4√(x(4-x)) √(x(4-x)) 4√(x(4-x))
= x3-3×2+4x-6
2√(x(4-x))
9.17. />
y’= x/2*√(6x-x2-8)+(x-3)(3-x) +1 = продолжение
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2√(6x-x2-8) 4√(x/2-1)√(1-x/2+1)
= 6×2-x3-8x-x2+9+2 = 5×2-x3-8x+10
2√(6x-x2-8) 4√((4-x)(x-2)) 2√(6x-x2-8)
9.18. />
arctg√x+1+x _ 1_
y’= 2(1+x)√x 2√x= arctg√x
x2x2
9.19. />
-arcsin√x+√(1-x)_
y’= √(1-x) √(x(1-x))= -arcsin√x+1 __ 1_
x2x2√(1-x) x2√x x√x
9.20. />
y’= x/2*√(5x-4-x2)+(2x-5)(5-2x)+9√3 = 16×2-16x-4×3-4×2+25+12√3√(x-1)
8√(5x-4-x2) 12√(3-x+1) 8√(5x-4-x2)
9.21. />
y’= 1 +5(x2+4)(2x(x2+4)-2x(x2+1))= x2+5x+4 _
1+x26(x2+1)(x2+4) (x2+1)(x2+4)
9.22. />
y’= √2(x-1)-√2(x-2) = 1 _
2(x-1)2√(1-(x-2)2/(2(x-1)2)) (x-1)√(x2-2)
9.23. />
y’= -x/√(1-x2) — arcsin√(1-x2)+2×2/√(1-1+x2)= -x/√(1-x2) — arcsin√(1-x2)+2x
9.24. />
y’= 1 +1 +16 = 1 +1 +4 _
2√x 6(1+x)√x 12(2+x)√x 2√x 6(1+x)√x 3(2+x)√x
9.25. />
√x-1 + √(1-x)
y’= (1-√x)2* 2√(1-x) 2√x = x-√x+1-x = 1 _
(1-√x)2+(1-x) (1-√x)22√(x(1-x))(1-√x)(1-√x+1+√x) 4√(x(1-x))
9.26. />
y’= (4x+6)arctg((x+1)/(x+2))+(2×2+6x+5)(x+2-x-1) – 1 =
(x+2)2(1+((x+1)/(x+2))2)
= (4x+6)arctg((x+1)/(x+2))
9.27. />
2√(1-4×2)+2x2_
y’= √(1-4×2)* arcsin2x + 2x _ 8x =
4(1-4×2) 2√(1-4×2) 8(1-4×2)
= 2(1-4×2)+2x2arcsin2x+x _ x = 1-3x2arcsin2x продолжение
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4(1-4×2) 1-4×21-4×22-8×2
9.28. />
y’= (4x-1)arctg((x2-1)/(x√3))+(2×2-x+1/2)(2×2√3-√3(x2-1))_ 3x2_ √3=
3×2(1+(x2-1)2/(3×2)) 2√3 2
= (4x-1)arctg((x2-1)/(x√3))+√3(2×2-x+1/2)(x2+1)_ 3x2_ √3
3×2+(x2-1)22√3 2
9.29. />
y’= (1+1/√x)arctg(√x/(√x+2))+x+2√x+2 _ 1 = (1+1/√x)arctg(√x/(√x+2))
√x((√x+2)2+x) 2√x
9.30. />
y’= 1-x arcsinx√2+√(1+2x-x2)(√2(1+x)-x√2)= 1-x arcsinx√2+√2_
√(1+2x-x2) 1+x (1+x)2√(1-2×2/(1+x)2) √(1+2x-x2) 1+x 1+x
9.31. />
y’= 4 = 1 _
4cos2(x/2)(4+(tg(x/2)+1)2) cos2(x/2)(4+(tg(x/2)+1)2)