Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm given implicit function $z=F(\frac{x}{z},\frac{y}{z})$. For $z(x,y)$ I want to show that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z.$$ I'm using implicit function theorem. My reasoning goes as follows. I compute matrix of linear derivative of $\textbf{F}(x,y,z)=F(\frac{x}{z},\frac{y}{z})-z$ with respect to $z$, assume it's not singular and reverse it. Next I compute linear derivative of $\textbf{F}$ with respect to $x,y$. Then I multiply the last matrix by the reverse of linear derivative of $\textbf{F}$ (from left). Finally I get matrix of linear derivative of the function $z(x,y)$ which should easily yield the result. By computation I get $$D_{z}\textbf{F}=\frac{\partial F}{ \partial u} \frac{\partial u} {\partial z} + \frac{\partial F}{ \partial t} \frac{\partial t} {\partial z}-1=-x\frac{\partial F}{ \partial u} \frac{1} {z^{2}} -y \frac{\partial F}{ \partial t} \frac{1} {z^{2}}-1,$$ so the reverse of $D_{z}\textbf{F}$ is $$[D_{z}\textbf{F}]^{-1} = (\frac{\partial F}{ \partial u} \frac{\partial u} {\partial z} + \frac{\partial F}{ \partial t} \frac{\partial t} {\partial z}-1)^{-1} $$and partial derivatives of $z(x,y)$ came up to be $$ \frac{\partial z}{ \partial x} = \frac{\frac{\partial F}{ \partial u} \frac{1} {z}}{[D_{x}\textbf{F}]^{-1}}, $$ and $$\frac{\partial z}{ \partial y} = \frac{\frac{\partial F}{ \partial t} \frac{1} {z}}{[D_{x}\textbf{F}]^{-1}}.$$ So it's easy to see that the equation does not hold. My question is: is my reasoning correct? And did I messed up something with partial derivatives?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.