Checking that the value $\phi'(1)$ agrees with the implicit function theorem lemma (implicit differentiation formula) where $\phi(x)$ was defined as
$$
\phi(x) =
\begin{cases} \phi_2(x), & x \in (-1,1) \\
1, & x = 1 \\
\phi_1(x), & x \in (1,\infty)\\
\end{cases}
$$
where $f(x,y) = 4y^3-3y-x = 0$
$\phi_1(x) =$ (all that stuff)
$\phi_2(x) = \cos(\frac{1}{3}\arccos(x)),\ x \in (-1,1)$
Rough solution: So taking the derivative:
$DF(x,y) = [-1, 12y^2 - 3]$
from implicit differentiation: $$\frac{\partial \phi}{\partial x}(1) = - \frac{\frac{\partial f}{\partial x}(1,1)}{\frac{\partial f}{…