Determine whether an implicit function $y=f(x)$ is given by the equation $F(x,y) = 0$ for some neighborhood around point $A$.

The heart of it is the fundamental theorem of implicit functions, which only provides sufficient conditions.
If all the following conditions are satisfied

  • Exists $\theta >0$ s.t $F$ is continuous in $U_\theta (A)$ and $F_y$ [the partial w.r.t $y$] is continuous in $U_\theta (A)$.
    • $F(A)=0$
    • $F_y(A) \neq 0$

then $F$ determines an implicit function $y=f(x)$ in $U_\delta (A)$ for some $\delta >0$.

  • If additionally, $F_x$ is continuous in $U_\theta (A)$ then for every $x\in (a-\delta,a+\delta)$ $$f'(x) = -\frac{F_x(x,f(x))}{F_y(x,f(x))}. $$ (i.e $f$ is continuously diff-ble)

Suppose now $F(x,y) := y^2x^{1/3}+\sin y$. It is then easy to verify the first three conditions, which will be sufficient for the existence of an implicit $y=f(x)$ in $U_\delta(A)$, where $A=(0,0)$. But the fourth condition is not met. We can't conclude that $f$ isn't differentiable at any point, can we? Suppose, I wanted to know what $f'(0)$ was, the formula given by theorem doesn't apply. So we have a new problem $$y^2x^{1/3}+\sin y=0\Longrightarrow y = ...? $$ Is there another way to either find the derivative or exclude its existence?

  • $\begingroup$ is here assumed to be that $y=y(x)$? $\endgroup$ – Dr. Sonnhard Graubner Mar 26 '17 at 13:10
  • $\begingroup$ @Dr.SonnhardGraubner Yes, the first three conditions of theorem guarantee that. One hope would be to make $y=y(x)$ i.e express it explicitly in terms of $x$, but that seems to be a pipedream. $\endgroup$ – Alvin Lepik Mar 26 '17 at 13:16

or you can write $$x=\left(-\frac{\sin(y)}{y^2}\right)^3$$ additionally we can get $y'$ as follows $$y'\left(2yx^{1/3}+\cos(y)\right)=-\frac{y^2}{3\sqrt[3]{x^2}}$$ if $$2yx^{1/3}+\cos(y)\ne 0$$ you can solve this for $y'$

  • $\begingroup$ I'm confused. Is it supposed to be the inverse of $y$? So I would get $y'(x) = \frac{1}{x'(y)}$? $\endgroup$ – Alvin Lepik Mar 26 '17 at 13:25
  • $\begingroup$ one moment please i will try to show that the inverse function exists $\endgroup$ – Dr. Sonnhard Graubner Mar 26 '17 at 13:27
  • $\begingroup$ what you have written is a possible solution $\endgroup$ – Dr. Sonnhard Graubner Mar 26 '17 at 13:27

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