Question on partial derivative with $w=w(x,y,z) and z=z(x,y)$

Question

I was concerning about $\dfrac{\partial w}{\partial x}$

I am thinking about using the chain rule as z depends on x and get $\dfrac{\partial w}{\partial x}$=$\dfrac{\partial w}{\partial x}$+$\dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial x}$

(ie if $w=xyz$ and $z=xy^2$ then $\dfrac{\partial w}{\partial x}=yz+xy \cdot y^2$)

but this seems to be wrong with $\dfrac{\partial w}{\partial x}$ on both sides

How should I do this? Also I would like to ask if we turn y to be subject in $z=z(x,y) $ is $\dfrac{\partial w}{\partial z}$ still the same? (ie $w=xyz$ and $z=xy^2$ vs $w=xyz$ and $y^2=x/z$)

Answer 1

The two $\dfrac{\partial w}{\partial x}$ terms stand for totally different things. For instance, if $w = x + y + z$ and $z = x^2$, then you have $w = x + y + x^2$.

If you regard $w(x,y,z) = x + y + z$ as a function of $x,y,z$ then $\dfrac{\partial w}{\partial x} = 1$.

If you regard $w(x,y) = x + y + x^2$ as a function of $x,y$ then $\dfrac{\partial w}{\partial x} = 1 + 2x$.

This is a shortcoming of the notation. The meaning is dependent on context.

Answer 2

You may want to write: $$\frac{\partial w(x,y,z(x))}{\partial x}=\frac{\partial w(x,y,z))}{\partial x}+\frac{\partial w(x,y,z))}{\partial z}\cdot\frac{d z(x))}{d x}$$

just letting clear that the left hand $w$ is a composite function.

Answer 3

I think the detailed would be: \begin{gather*} \frac{\partial w(x,y,z(x))}{\partial x}=\frac{\partial w(x,y,\gamma)}{\partial x}\Bigg|_{\gamma=z(x）}+\frac{\partial w(x,y,\gamma)}{\partial \gamma}\Bigg|_{\gamma=z(x）}\cdot\frac{d z(x)}{d x}. \end{gather*} Thus we see that the left side hand says something concerning differentiation after composition, while the right side hand composition after differentiation.