How do you implicitly differentiate # y-(3y-x)^2-1/y^2=1-x#?

Question

916 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-03T17:19:05

#dy/dx=(-1-6y+2x)/(1-18y+6x+1/y^3 )#

#y-(3y-x)^2-1/y^2=1-x#

#y-(3y-x)^2-y^-2=1-x#

#dy/dx+2(3y-x)-6(3y-x)dy/dx+1/y^3 dy/dx=-1#

#dy/dx-6(3y-x)dy/dx+1/y^3 dy/dx=-1-2(3y-x)#

#dy/dx(1-6(3y-x)+1/y^3 )=-1-2(3y-x)#

#dy/dx(1-18y+6x+1/y^3 )=-1-6y+2x#

#dy/dx=(-1-6y+2x)/(1-18y+6x+1/y^3 )#