Question

If x^3+2x^2y-4y=7, then when x=1, dy/dx=

Answer 1

Answer:

dy/dx when x = 1 is (4y + 3) / 2.

Step-by-step explanation:

x^3 + 2x^2y - 4y=7

Using implicit differentiation:

3x^2 + 4xy + 2x^2.dy/dx - 4 dy/dx = 0

2x^2.dy/dx - 4dy/dx = -3x^2 - 4xy

dy/dx =  (-3x^2 - 4xy ) / (2x^2 - 4)

When x = 1:

dy/dx =  (-3 - 4y) / -2  

= -1(4y + 3) / -2

(4y + 3)/ 2 (answer).