Use implicit differentiation to find the slope of the tangent line to the curve:
-2x^2 + 3xy + 4y^3 = -5 at the point (3,1).
2007-11-10 06:19:30 · 4 answers · asked by m_mays89 2 in Science & Mathematics ➔ Mathematics
-4x + 3x dy/dx + 3y + 12y^2 dy/dx = 0
(3x+12y^2)dy/dx = 4x - 3y
sub (3,1)
(9+12)dy/dx = 12-3 = 9
dy/dx = 9/21 = 3/7
2007-11-10 06:26:47 · answer #1 · answered by norman 7 · 0⤊ 0⤋
-4x+3y+3xy´+12y^2y´=0
y´(3x+12y^2)=4x-3y
at (3.1) y´(21)=9 so y´= slope = 3/7
2007-11-10 14:29:14 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
-4x + 3x dy/dx + 3y + 12y^2 dy/dx = 0
(3x + 12y^2) dy/dx = 4x - 3y
dy/dx = (4x - 3y)/ (3x + 12y^2)
at (3,1).
dy/dx = (12 -3)/(9+12) = 9/21 = 3/7
2007-11-10 14:38:10 · answer #3 · answered by Anonymous · 0⤊ 0⤋
-4x +3xy'+3y +12y^2y' =0
3y'(x+4y^2) = 4x-3y
y' = (4x-3y)/3(x+4y^2)
y' = (4*3-3*1)/(3+4) = 3/7
2007-11-10 14:28:21 · answer #4 · answered by ironduke8159 7 · 0⤊ 1⤋