An inplicit differentiation question

Question

Given that, $$z=f(\frac{ny-mz}{nx-lz})$$ prove that , $$(nx-lz)\frac{\partial z}{\partial x}+(ny-mz)\frac{\partial z}{\partial y}=0 $$. where $l,m,n $ are constants . In my notebook , it is given that $$\frac{\partial} {\partial x}\frac{ny-mz}{nx-lz} = (ny-mz)*\frac {-n}{(nx-lz)^2}$$ which seems incorrect to me as the numerator contains $z$ which itself depends on $x$.My question is , how do i then compute $\frac{\partial} {\partial x}\frac{ny-mz}{nx-lz}$ ?

Thanks for reading.

Answer 1

The assertion is true provided the LHS of your equation is zero, not $z$. Suppose $$f\left(\frac{ny-mz}{nx-lz}\right)=0\tag1$$ where $z$ is a function of $x$ and $y$. Writing $$u(x,y,z):= \frac{ny-mz}{nx-lz},\tag2$$ we differentiate (1) with respect to $x$ and $y$ using the chain rule to obtain $$ \def\pard#1#2{\frac{\partial{#1}}{\partial{#2}}} f'(u)\left(\pard ux+\pard uz\frac{\partial z}{\partial x}\right)=0\qquad\text{and}\qquad f'(u)\left(\pard uy + \pard uz\pard zy\right)=0 $$ respectively. (Note we are treating $u$ as a function of independent $x$, $y$, $z$ when computing partial derivatives.) The expressions in parentheses equal zero, so eliminate $\partial u/\partial z$ and rearrange to get $$\pard uy\pard zx = \pard ux\pard zy.$$ The result follows from differentiating (2) to obtain $$\pard uy=\frac {n(nx-lz)}{(nx-lz)^2}\qquad\text{and}\qquad\pard ux=\frac{-n(ny-mz)}{(nx-lz)^2},$$ plugging in, and simplifying.

Answer 2

Here's the correct approach. You have a function $F(x,y,z)=0$ which defines $z$ implicitly as a function of $x,y$. In your case $F(x,y,z) = f(\dfrac{ny-mz}{nx-lz}) - z$.

Assuming this equation defines $z$ as a function of $x,y$, differentiate the function $\Phi(x,y)=F(x,y,z(x,y))$, which is identically $0$, with respect to $x$ and $y$ using the chain rule.