# How to evaluate derivative for implicit

If $$x = \frac{1}{2}y(y^2-3)$$, how do I use Mathematica to evaluate $$\frac{\text{d}y}{\text{d}x}$$ at the point where $$y=0$$? Doing f[y_] = 1/2*y*(y^2 - 3) and then D[y, f[y]] /. y -> 0 gives the errors "$$\frac{1}{2}y(-3+y^2)$$ is not a valid variable" and "$$0$$ is not a valid variable".

• Maybe you mean: D[f[y], y] /. y -> 0 ? Oct 26, 2019 at 23:24
• But as $x=f(y)$, that would give me $\frac{\text{d}x}{\text{d}y}$, whereas I want $\frac{\text{d}y}{\text{d}x}$. Oct 26, 2019 at 23:25
• So then maybe you are looking for: D[InverseFunction[f][y], y] /. y -> 0? Oct 26, 2019 at 23:43
• Yes, that works in this case. Oct 27, 2019 at 0:05
• Oct 27, 2019 at 2:32

## 3 Answers

This is basically the same approach as coolwaters, just packaged up differently. You can use Dt:

Dt[x == 1/2 y (y^2-3), x]


1 == y^2 Dt[y, x] + 1/2 (-3 + y^2) Dt[y, x]

Using Solve, and then replacing y with 0:

Dt[y, x] /. First @ Solve[Dt[x == 1/2 y (y^2-3), x], Dt[y, x]] /. y->0


-2/3

One approach is to find an explicit representation for the inverse function, and then take the derivative:

D[InverseFunction[f][y], y] /. y -> 0


Consider y as a function of x:

d = D[x == 1/2 y[x] (y[x]^2 - 3), x]
(* 1 == y[x]^2 y'[x] + 1/2 (-3 + y[x]^2) y'[x] *)


Replace y[x] by 0:

d /. y[x] -> 0
(* 1 == -(3/2) y'[x] *)


Which solves as y'[x] -> -(2/3). Note this is only valid if the original equation has a solution after letting y == 0.