Matematica

Tarea de Matemática II.

Hallar dy/dx por derivación implícita si :

X2y3-2xy+y2=x3+1

X2.3y2dy/dx+y3.2x-[2xdy/dx+y.2]+2ydy/dx=3x2

3x2y2dy/dx+2xy3-2xdy/dx-2y+2ydy/dx=3x2

dy/dx [3x2y2-2x+2y] = 3x2-2xy3+2y

dy/dx = (〖3x〗^2-〖2xy〗^3+2y )/(〖3x〗^2 y^2-2x+2y)

2xy4+5y2-6x = y+6

2x.4y3dy/dx+y4.2+10dy/dx -6= ydy/dx

8xy3dy/dx+2y4+10dy/dx-6= ydy/dx

dy/dx [8xy3+10-1]= -2y4+6

dy/dx= (〖-2y〗^4+6 )/(8xy^3+10-1)

4x3y5-6xy+x4=x2

4x3.5y4dy/dx+y5.12x2-[6xdy/dx+y.6]+4x3=2x

20x3y4dy/dx+12x2y5-6xdy/dx -6y+4x3=2x

dy/dx [20x3y4-6x]=2x-12x2y5+6y-4x3

dy/dx= (〖2x-12x^2 y〗^5+6y- 〖4x〗^3)/(20x^3 y^4-6x)

5y+6x3-xy=y2+x

5dy/dx+18x2-[xdy/dx+y.1]=2ydy/dx +1

5dy/dx+18x2- xdy/dx-y=2ydy/dx +1

dy/dx [5-x-2y]=1-18x2+y

dy/dx = (〖1-18x^2〗^ + y)/(5-x-2y)

Hallar la diferencia dy si:

Y=4x3+2x2+6x+10

dy= (4x3+2x2+6x+10) (12x2+4x+6)dx

Y = cos (3x2)

dy = -sen (3x2) (6x) dx

Y = sen (6x3)

dy = cos (6x3) (18x2) dx

Y = 5x4+6x3+2y2-2y+1

dy = (5x4+6x3+2y2-2y+1) (20x3+18x2+4y-2) dx

Y = -sen (x5)

dy = cos (x5) (5x4) dx

Hallar máximos y mínimos para las siguientes funciones:

Y= 2x3+x2 S = [ -1, 6 ]

C1=-1 C2=6 C3=0 C4= -1/3

6X2+2X

Igualando a cero

6x2+2x=0

2x[3x+1]=0

X=[3x+1]= 0/2 =0

X=[3x+1]= 0

X=0 ^ 3x+1=0 3x=0-1 x=0 (-1)/3 X= (-1)/3

f(-1)=2(-1)3+(-1)2

f(-1)=2(-1)+1

f(-1)=-2+1 =-1

f(6)= 2(6)3+(6)2

f(6)=2(216)+36

f(6)=468

f(0)=2(0)3+(0)2= 0

f((-1)/3 )= 2((-1)/3 )3+((-1)/3 )2

f((-1)/3 )=2((-1)/27)+ 1/9

f((-1)/3 )= (-2)/27 + 1/9 = -0.074+0.111 = 0.037

f(x) tiene un máximo en f(c)=468 que lo alcanza cuando C =6

f(x) tiene un minimo en f(c)= -1 que lo alcanza cuando C = -1

Y= x3-2x2+6 S= [ 0, 5 ]

C1= 0 C2 = 5 C3 = 0 C4 = 4/3

F'(x) = 3x2 – 4x

Igualando a cero

3x2 – 4x = 0

X (3x – 4)= 0

x = 0 ^ 3x – 4 = 0

3x = 0 + 4

x = 4/3

F (0) = 03 – 2(0)2 + 6 = 6

F (5) = 53 – 2(5)2 + 6

=125 – 2 (25) + 6

= 125 – 50 + 6

= 81

F (0) = 03 – 2(0)2 + 6 = 6

F (4/3) = (4/3)3 – 2(4/3)2

...