Question

Q30. If y^3 + xy^2-2x = 0 defines y implicitly as a function of x, then the value of
dy/dx
at the point (4, -2) is
(A) - 1/2
(B)- 1/8
(C) 1/4
(D) 1/2

Answer 1

Okay, using implicit differentiation we get D(y^3 + xy^2 - 2x) = 3y^2y' + y^2 + 2xyy' - 2 = 0. We can see then that y' = (-y^2 +2) / (3y^2 + 2xy). Then we evaluate this at (4,-2) to get that y' = -2 / -4 = 1/2.