Find ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$

Question

I'm currently learning how to use partial differentiation. Can someone give me a hint?

find ${\partial z\over \partial x}$ and ${\partial z \over \partial y}$ for $xy + yz = xz$

Answer 1

Assuming that $x$ and $y$ are independent variables, differentiating both sides with respect to $x$ using product rule, we get, \begin{equation*} y+y\dfrac{\partial z}{\partial x} = z+x\dfrac{\partial z}{\partial x} \end{equation*} Solving this we get the desired result for $\dfrac{\partial z}{\partial x}$. Similarly for $y$.

You can also check that answer obtained this way is correct answer by explicit differentiation: \begin{align} &z=\dfrac{xy}{x-y}\\ \implies &\dfrac{\partial z}{\partial x} = -\dfrac{y^2}{(x-y)^2} \end{align}

And answer obtained above is \begin{equation} \dfrac{\partial z}{\partial x} = \dfrac{y-z}{x-y} \end{equation}

Substituting $z=\dfrac{xy}{x-y}$ in above equation, we get, \begin{equation} \dfrac{\partial z}{\partial x} = -\dfrac{y^2}{(x-y)^2} \end{equation}

Answer 2

$\def\p#1#2{\frac{\partial #1}{\partial #2}}$Calculate the differential of the entire equation $$\eqalign{ 0 &= d(xy + yz - xz) \\ &= d(xy) + d(yz) - d(xz) \\ &= (x\,dy + y\,dx) + (y\,dz + z\,dy) - (x\,dz + z\,dx) \\ (x-y)\,dz &= (y-z)\,dx + (x+z)\,dy \\ }$$ Holding $x$ constant (i.e. setting $\,dx=0$) yields $$\eqalign{ (x-y)\,dz &= (x+z)\,dy \quad\implies\quad \p{z}{y} = \frac{x+z}{x-y} \\ }$$ While holding $y$ constant ($dy=0$) yields $$\eqalign{ (x-y)\,dz &= (y-z)\,dx \quad\implies\quad \p{z}{x} = \frac{y-z}{x-y} \\ }$$