Como le pongo los puntos a la i

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  • Publicado : 10 de noviembre de 2011
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SOLUCION

1. y = 5x4 + 8x3 – 2x2 + 5x – 4
y’ = 20x + 4x – 4x + 5

2. y = (x2 + 3)2
y’ = 4x(x2 + 3)

3. y = 125x3 + 11x2 – 25x + 30
y’ = 375x2 + 22x – 254. y = (x + 1) (x2 + 1)2
y’ = (x + 1) (4x3 +4x)
y’ = 1(4x3 + 4x) (12x2 + 4)(x + 1)
y’ = 4x3 + 4x + 12x3 +12x2 + 4x + 4
y’= 16x3 +12x2 +8x +4

5. y=[pic]
y’ = 2x (x2 – 1) – 2x(x2)
(x2 – 1)2
y’ = 2x3 – 2x - 2x3
(x2 – 1)2

y’ = - 2x
(x2 – 1)2

y’ =- 2x
4x2 – 2x + 1

y’ = - 2 (4x2-2x + 1) – (8x – 2)(-2x)
(4x2 – 2x + 1)2

y’ = - 8x2 + 4x - 1 + 16x2 – 4x(4x2 – 2x + 1)2

y’ = -8x2 – 1
(4x2 – 2x + 1)2

6. y = x - 1
x + 1
y’= 1(x+1) – 1(x-1)
(x + 1)2
y’=x+1 – x + 1
(x + 1)2

y’= 2 = 2(x + 1)-2
(x + 1)2
y’’ = -4(x + 1)-3 (1)
y’’ = -4(x + 1)-3
y’’ = 12(x + 1)-4 (1)
y’’= 12(x + 1)-4

7. yIV si y = 1
4x – 2
y’ = 0(4x – 2) -4 (1)
(4x – 2)2

y’= -4(4x – 2)-2
y’’ = 8(4x – 2)-3 (4)y’’ = 32(4x – 2)-3
y’’’ = -96(4x – 2)-4 (4)
y’’’ = -384(4x – 2)-4
yIV = 1536(4x – 2)-5 (4)
yIV = 6149(4x – 2)-5

8. s = 10t + 25t2
s’ = 10 + 50t=> Velocidad
s’’ = 50 => Aceleración

9. s = t(3 – t)

s’= 1(3 – t) + (-1)(t)
s’= 3 – t – t
s’ = 3 – 2t
s’’ = -2
s’’’ = 0

10.
a.C(x) = 300 + 50 x – 0,1x2 + 0,003x2
C’(x) = 50 – 0,2x + 0,006x
C’(x) = 50 – 0,194x

b. C(x) = 45x2 + 30x – 10x + 5
C’(x) = 90x + 30 – 10
C’(x) = 90x +...
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