If we have known $f(x,y)=0$ defines an implicit function $y=g(x)$.
Can we derive $g$ is differentiable from $f$ is differentiable with respect to $x$?
2026-04-01 15:56:25.1775058985
Differentiability of Implicit Function
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No, $f(x,y)=y/(2+sign(y))-x$ defines the implicit function $y(x) = x(2+sign(x))$, which is not differentiable at $x=0$ even though $f(x,y)$ is $C^1$ in $x$.