Question

A função y = f(x) ́e dada implicitamente pela equaçao 3y^2 +2xy -x^2=3. Determine y' (1/2)

Answer 1

Resposta:

3y^2 +2xy -x^2=3

3*2*y*dy/dx+2y +2x*dy/dx -2x=0

dy/dx * (6y+2x)  +2y-2x=0

dy/dx = 2*(x-y)/(6y+2x) =(x-y)*(3y+x)

para x=1/2

==>3y^2 +2*(1/2)y -(1/2)^2=3

==>3y^2 +y -1/4=3

3y²+y-13/4=0

y'=[-1+√(1+39)]/6=-1/6 +√10/3

y''=[-1+√(1+39)]/6=-1/6 -√10/3

y'(1/2 , -1/6 +√10/3)

dy/dx =(1/2-(-1/6 +√10/3))/(3*(-1/6 +√10/3)-1/2)

y'(1/2 , -1/6 -√10/3)

dy/dx = (1/2-(-1/6 -√10/3))/(3*(-1/6 -√10/3)-1/2)