Calculus I - Implicit Differentiation
tutorial.math.lamar.edu › Classes › CalcIMay 30, 2018 · 2 x + 2 [ y ( x)] 1 y ′ ( x) = 0 2 x + 2 [ y ( x)] 1 y ′ ( x) = 0. At this point we can drop the ( x) ( x) part as it was only in the problem to help with the differentiation process. The final step is to simply solve the resulting equation for y ′ y ′. 2 x + 2 y y ′ = 0 y ′ = − x y 2 x + 2 y y ′ = 0 y ′ = − x y.