For a second degree function as $y^2 = x$, from implicit differentiation we obtain: $\frac{dy}{dx} =\frac{ 1}{2y}$ , there are two possible values of $y$; so the curves is either: $ y' = \frac{1}{2\sqrt{x}}$ or $y' = \frac{-1}{2\sqrt{}x}.$

Now, these are roots that define the curve of derivatives.

For a third degree function such as $y^3 + x^3 + 9xy = 0$, how would we even plot the three graphs of the function for which the respective graphs of derivatives exist, because only one root is real and two roots are the complex conjugates. Do we only consider the real value? If so, what are the uses of the two complex conjugates for defining the derivative?

We obtain a tangent,after differentiation, but to which curve?



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