Turunan pertama dari x^3-3x^2 y+ y^2 = 0
Posted by andi telaumbanua on Feb 13, 2018 in Matematika |
Tentukan turunan pertama dari bentuk implisit x^3-3x^2 y+ y^2=0
Jawab :
Turunan bentuk implisit
(d(x^3 – 3x^2 y + y^2))/dx = (d(0))/dx
d(x^3 )/dx – d(3x^2 y)/dx + d(y^2 )/dx = 0
3x^2 – [(d(3x^2 )/dx)(y) + (3x^2 )(dy/dx) ]+(d(y^2 )/dy)(dy/dx) = 0
3x^2 – [6xy+(3x^2 )(dy/dx) ]+ 2y (dy/dx)=0
3x^2- 6xy – 3x^2 dy/dx + 2y dy/dx = 0
2y dy/dx – 3x^2 dy/dx = 6xy – 3x^2
dy/dx (2y – 3x^2 )= 6xy – 3x^2
dy/dx = (6xy – 3x^2)/(2y – 3x^2 )
