 equissupnica7

2022-07-16

There is no trouble when I have to differentiate a function $f\left(x,y\right)$ where $y=g\left(x\right)$ using Implicit Function Theorem. It's easy to get ${g}_{xx}$ and ${g}_{yy}$. For instance, the formula for ${g}_{xx}=-\frac{{f}_{xx}{F}_{y}^{2}-\left({F}_{xy}+{F}_{yx}\right){F}_{x}{F}_{y}+{F}_{yy}{F}_{x}^{2}}{{F}_{y}^{3}}$ knowing that ${g}^{\prime }\left(x\right)=-\frac{{F}_{x}}{{F}_{y}}$ whenever ${F}_{y}\ne 0$.
Now, i'm in the situation $z+x+\left(y+z{\right)}^{4}=0$ where it defines an implicit function $z=f\left(x,y\right)$.
I got ${f}_{x}$ and ${f}_{y}$ using the theorem. However, I can't get ${f}_{xx}$ and ${f}_{yy}$ from there. I know I shoud apply the chain rule but I get lost whenever trying to get a "general formula" for them. sviudes7w

Expert

From $F\left(x,y,z\left(x,y\right)\right)=0$,
$\begin{array}{rl}& \frac{\mathrm{\partial }}{\mathrm{\partial }x}F={F}_{x}+{F}_{z}{z}_{x}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{z}_{x}=-\frac{{F}_{x}}{{F}_{z}}\\ & \frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{x}^{2}}F={F}_{xx}+{F}_{xz}{z}_{x}+{z}_{x}\left({F}_{zx}+{F}_{zz}{z}_{x}\right)+{F}_{z}{z}_{xx}=0\end{array}$
Plugging in ${z}_{x}$ and rearranging, it should be clear how to solve for ${z}_{xx}$. A similar method holds for $y$.

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