Calculating the second order partial derivatives of $z(x,y.)$.

Question

If we know that $z_{x}^{'} = \frac{yz}{z^{2} - xy },$, how can I calculate the second order partial derivative with respect to x, knowing that the final answer should be (as given at the back of the book) $z_{xx}^{''} = \frac{2xy^{3}z}{(z^{2} - xy)^3 },$. could anyone explain to me how to do it? I do not know from where the power 3 in the denominator came.

Answer 1

If we know that $z_{x}^{'} = \frac{yz}{z^{2} - xy }$, how can I calculate the second order partial derivative with respect to x

Take the partial derivative with respect to $x$: use the quotient rule and don't forget that $z$ is a function of $x$ (and $y$). This will then contain the first partial derivative $z'_x$ again, but you can substitute that with: $$z_{x}^{'} = \frac{yz}{z^{2} - xy }\tag{$*$}$$ and then simplify.

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Here's a start: $$\begin{align}z_{xx}^{''}=\frac{\partial}{\partial x}z_{x}^{'} & =\frac{\partial}{\partial x}\left(\frac{yz}{z^{2} - xy }\right) \\[6pt] & =\frac{\left(z^{2} - xy\right)\color{red}{\frac{\partial}{\partial x}\left(yz\right)}-\left(yz\right)\color{blue}{\frac{\partial}{\partial x}\left(z^{2} - xy\right)}}{\left(z^{2} - xy\right)^2 } \\ & = \ldots \end{align}$$ Now $\color{red}{\frac{\partial}{\partial x}\left(yz\right)}=yz_{x}^{'}$ and use $(*)$ for $z_{x}^{'}$ and don't forget the chain rule for $\color{blue}{\frac{\partial}{\partial x}\left(z^{2} - xy\right)}$ and you'll be able to use $(*)$ once again to eliminate $z_{x}^{'}$.