How do you implicitly differentiate # x^3 - 3xy + 2y^3 = 3#?

1 Answer

#dy/dx=(y-x^2)/(2y^2-x)#

Explanation:

We assume that y is a function of x, ie #y=f(x)# and then differentiate each side of the equation with respect to x, then re-arrange and solve for #dy/dx#.

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

#therefore 3x^2-3xdy/dx-3y+6y^2dy/dx=0#. (Used product rule and power rule)

#therefore dy/dx(-3x+6y^2)=3y-3x^2#. (Took out common factor)

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

#therefore dy/dx=(y-x^2)/(2y^2-x)# (Divide all by 3).