Find the 1st and 2nd derivative of 3xy=4x+y^2

Question

I can't seem to find the answer help please!

 

Answer 1

Given function 3xy = 4x+y^2

Apply derivative to each sides.

d/dx(3xy) = d/dx(4x+y^2)

3y+3xdy/dx = 4+2ydy/dx ----> (1)

dy/dx(3x-2y) = 4-3y

1st derivative is dy/dx = (4-3y)/(3x-2y) ----> (2)

3y+3xdy/dx = 4+2ydy/dx

Again derivative apply to above equation.

 3dy/dx+3dy/dx+3xd^2y/dx^2 = 2(dy/dx)^2+2yd^2y/dx^2

6dy/dx+3xd^2y/dx^2-2yd^2y/dx^2 = 2(dy/dx)^2

d^2y/dx^2(3x-2y) = 2(dy/dx)^2-6dy/dx

d^2y/dx^2(3x-2y) = 2dy/dx((dy/dx)-3)

Substitute dy/dx value from (2) in above equation.

d^2y/dx^2(3x-2y) = 2(4-3y)/(3x-2y)[(4-3y)/(3x-2y)-3]

d^2y/dx^2(3x-2y) = 2(4-3y)/(3x-2y)[(4-3y)-3(3x-2y)]/(3x-2y)

d^2y/dx^2(3x-2y) = (8-12y)/(3x-2y)[4-3y-9x+6y]/(3x-2y)

Second derivative is d^2y/dx^2 = (8-12y)(-9x+3y+4)/(3x-2y)^2