Finding derivative of implicit function at a point

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?

• is here assumed to be that $y=y(x)$? – Dr. Sonnhard Graubner Mar 26 '17 at 13:10
• @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. – 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 $y'$
• I'm confused. Is it supposed to be the inverse of $y$? So I would get $y'(x) = \frac{1}{x'(y)}$? – Alvin Lepik Mar 26 '17 at 13:25