How do you do implicit differentiation of #y^3+x^2y+x^2-3y^2#?

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Answer 1 · 2015-05-21T03:02:32

Each #y# is some (unknown) function of #x#, so

to differentiate #y^3# and #-3y^2# we will use the power rule and the chain rule (a combination sometimes called 'the generalized power rule')

To differentiate #x^2y# we'll need the product rule.
(I use the order: #(fg)' = f'g+fg'# and I'll put the product in brackets #[ . . ]#)

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

#color(white)"sssssss"# #= 3y^2 dy/dx +[ 2xy+x^2 dy/dx] +2x -6y dy/dx#

#color(white)"sssssss"# #= (2xy +2x) + (3y^2 +x^2 -6y )dy/dx#

That's all we can do. If we had an equation, we could solve for #dy/dx#.