Let 5x^2 - 3xy + 58 = 2y^3

Question

Use implicit differentiation to find y' for the above equation.

The first thing I did was move 2y^3 over to that everything is equal to 0. With the -3xy I used the product rule and expanded that.
After solving I got....

10x(3 * y + 3x * 1dy/dx)- 6y^2dydx= 0

= 10x-3y+3x*(1*dy/dx)-(6y^2*dydx)=0
Pulled out -6y^2*dy/dx (13x-3x)=0
Moved everything over and got...
-6y^2*dy/dx = 13x+3y
Moved -6y^2 over by dividing and the final answer I got was..

dy/dx = (-13x+3y)/(-6y^2)

Im searching online and its saying that I have the wrong answer.

Answer 1

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

(6y^2+3x)dy/dx = 10 x -3xy

dy/dx = (10x-3xy)/(6y^2+3x)

Answer 2

Just do it term by term the way it stands,
the only one to be careful with is the -3xy, since you have to use a product rule

5x^2 - 3xy + 58 = 2y^3
10x - 3x dy/dx -3y + 0 = 6y^2 dy/dx
10x - 3y = 6y^2 dy/dx + 3x dy/dx
factor out the dy/dx and just flip the equation
dy/dx(6y^2 + 3x) = 10x - 3y
dy/dx = (10x - 3y) / (6y^2 + 3x)

check:
http://www.wolframalpha.com/input/?i=find+dy%2Fdx+for+5x%5E2+-+3xy+%2B+58+%3D+2y%5E3

Answer 3

5x^2 - 3xy + 58 = 2y^3
10x-3y-3xy'=6y^2 y'

y'(6y^2+3x)=10x-3y

y'=(10x-3y)/(6y^2+3x)

check my work.

Answer 4

geepers, we all at at the same time got the same answer.

Answer 5

Nah, I carried an incorrect extra x

Answer 6

THANK YOU!!!!