Prove that $xz^3-yz=x$ is a function of $z$ at $(1,0,1)$

Question

Prove that $xz^3-yz=x$ is a function of z at (1,0,1).

Find $\frac{\partial \varphi}{\partial x}(1,0)$

Let the function be $f(x,y,z)=xz^3-yz-x$, by the implicit function theorem I found that:

• $f(x,y,z)$ is a $C^{\infty}$ function.
• $f(1,0,1)=0$
• $\frac{\partial f}{\partial z} (1,0,1) = 3 \neq 0$

So, $f(x,y,z)$ defines $z$ as a function of $(x,y)$ at $(1,0,1)$.

Now, when finding $\frac{\partial \varphi}{\partial x}(1,0)$ I get two different answers:

1. When using the formula: $\frac{\partial \varphi}{\partial x}(1,0)=-\frac{f_x(1,0,1)}{f_z(1,0,1)}$

$f_x(x,y,z)=z^3-1 \Rightarrow f_x(1,0,1)=0$

$\Rightarrow \frac{\partial \varphi}{\partial x}(1,0) = 0$

2. When not using the formula:

$xz^3-yz-x=0$, now let $z=\varphi(x,y)$

So, $f(x,y,\varphi(x,y))=x\varphi^3(x,y)-y\varphi(x,y)-x$

$\Rightarrow \frac{\partial \varphi}{\partial x} (x,y)$ $=\varphi^{3}(x,y)+3x\varphi^2(x,y)+\varphi'(x,y)x-y\varphi'(x,y)-1$

Now solving for $\varphi'(x,y) \Rightarrow$

$\varphi '(x,y)= \frac{1-3x\varphi^2(x,y)-\varphi^3(x,y)}{x-y} \Rightarrow \varphi'(1,0) = 1-3\varphi^2(x,y)-\varphi^3(x,y)$

Now, because $\varphi(x,y)=z=1 \Rightarrow \varphi'(1,0)=-3$

I've already checked for algebraic errors, but I don't seem to find one.

What's the correct method to solving this kind of problems?

Answer 1

Note that $f(x,y,\varphi(x,y))=0=x\varphi^3(x,y)-y\varphi(x,y)-x$ in a neighborhood of $(1,0,1)$. To arrive at the formula naturally, take partial $x$ of both sides.

Then $0=\varphi^3(x,y)+3x\varphi^2(x,y)\varphi_x-y\varphi_x-1=\varphi_x[3x\varphi^2-y]+\varphi^3-1$. Therefore, $$ \varphi_x=-\frac{(\varphi^3-1)}{3x\varphi^2-y} $$ Then you are done by realizing $\varphi(1,0)=1$.

Answer 2

The first method is correct.

In the second method, you tried to calculate $\frac{\mathrm d f}{\mathrm d x}$ instead of $\frac{\partial \varphi}{\partial x}$ in the first step. By the way: Trying to calculate the derivative by using the explicit form of the equation won't work in most cases, as the Implicit Function Theorem is often applied in cases where there is no analytic solution for $\varphi$ in this case.