Implicit differantiation

Question

Let z and w be differantiable functions of x and y and they satisfy the following equations.

$$ xw^3+yz^2+z^3=-1$$ $$ zw^3-xz^3+y^2w=1$$ Find $\frac{\partial z}{\partial x}$ and its value at $(x,y,z,w)=(1,-1,-1,1)$.

I don't know what to do with two implicit functions. So I thought maybe I can write $w$ in the first equation as a function of x,y and z such as : $$ w = \left( -\frac{1+yz^2+z^3}{x} \right) ^{1/3}$$ Then I can put it into the second equation: $$ z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3}=1$$ And here I can calculate the partial derivative as following: $$F(x,y,z) = z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3} $$ $$\frac{\partial z}{\partial x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} $$ Then I found the result as: $$ \frac{\partial z}{\partial x} = \frac{\frac{z(1+yz^2+z^3)}{x^2}-z^3+\frac{y^2(1+yz^2+z^3)^{1/3}}{3x^{4/3}}}{\frac{-(4z^3+3yz^2+1)}{x}-3xz^2-\frac{y^2\left(\frac{z(3z+2y)}{3(z^3+yz^2+1)^{2/3}} \right)}{3x^{4/3}}}$$

But I'm guessing that I missed something here by eliminating $w$. Could you please give me your opinions? Thank you.

Answer 1

Indeed, you don't need to eliminate $w$.

Differentiating the first equation with respect to $x$ gives you

$$ w^3 + 3 xw^2 \frac{\partial w}{\partial x} + 2 y z\frac{\partial z}{\partial x} + 3 z^2 \frac{\partial z}{\partial x}=0$$ and with the replacement $(x,y,z,w)=(1,-1,-1,1)$:

$$ 1 + 3 \frac{\partial w}{\partial x} + 5 \frac{\partial z}{\partial x}=0$$

Proceeding the same way by differentiating the second equation with respect to $x$, you get another linear equation linking $\frac{\partial z}{\partial x}$ and $\frac{\partial w}{\partial x}$.

You finally have a linear system of two equations linking $\frac{\partial z}{\partial x}$ and $\frac{\partial w}{\partial x}$. Solving it you get the desired $\frac{\partial z}{\partial x}$.

Answer 2

Since you have \begin{align} &x w^3 + y z^2 + z^3 = -1 \\ &z w^3 - x z^3 + y^2 w = 1, \end{align}

take the derivative with respect to $x$ for both equation, you have \begin{align} &x (3 w^2 \frac{\partial w}{\partial x}) + w^3 + y (2 z \frac{\partial z}{\partial x}) + 3 z^2 \frac{\partial z}{\partial x} = 0 \\ &z (3 w^2 \frac{\partial w}{\partial x}) + (\frac{\partial z}{\partial x}) w^3 - x (3 z^2 \frac{\partial z}{\partial x}) - z^3 + y^2 \frac{\partial w}{\partial x} = 0, \end{align}

then by substituting $(x, y, z, w) = (1, -1, -1, 1)$, you have \begin{align} & 3 \frac{\partial w}{\partial x} + 1 + 2 \frac{\partial z}{\partial x} + 3 \frac{\partial z}{\partial x} = 0 \\ -&3 \frac{\partial w}{\partial x} + \frac{\partial z}{\partial x} - 3 \frac{\partial z}{\partial x} + 1 + \frac{\partial w}{\partial x} = 0. \end{align}

Obviously, the solution for $\frac{\partial z}{\partial x}$ is $$ \frac{\partial z}{\partial x}|_{(x,y,z,w)=(1,-1,-1,1)} = -\frac{5}{4} $$