Chain Rule for Multivariable Calculus $dw/dz$ vs $(dw/dz)_y$

Question

I'm a high school math teacher teaching Multivariable Calculus. I posed the following question on a recent quiz that I found on MIT Open Courseware: Suppose that $x^2y+xz^2=5$ and let $w=x^3y$. Express $dw/dz$ as a function of x, y, z.

The original question asked students to find $(dw/dz)_y$. Does the question make sense the way I posed it? If not, what does the y-subscript mean? As a teacher, I liked the question because it combines explicit and implicit differentiation, but I feel like I made an error.

Answer 1

You need to make clear which variables are related to which. I'm assuming you want $z = z(x,y) $? Or $x = x(z), \ y = y(z)$?

Second, the notation $dw/dz$ implies treating $w$ as a single-variable function, which itself has two parameters $w = w(z) = w(z(x,y))$? Or do you want something like $w = w(x,y) = w(x(z),y(z))$ which is a function of two variables, and one parameter?

If it is the second case, then $$ \frac{dw}{dz} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial z} $$

If it's the first case, you'll have two simultaneous equations $$ \frac{\partial w}{\partial x} = \frac{dw}{dz}\frac{\partial z}{\partial x}, \ \frac{\partial w}{\partial y} = \frac{dw}{dz}\frac{\partial z}{\partial y} $$

Lastly, the subscript notation probably refers to the partial derivative $$ \left(\frac{dw}{dz} \right)_y = \frac{\partial}{\partial y} \left(\frac{dw}{dz} \right) $$

Answer 2

It makes sense, but would be confusing to HS students. If we know $w$ is a function of more that one variable then $\frac{d (w)}{dz}$ means that $x,y$ are functions of $z$ (maybe other variables too) i.e we can write this as $w = w(x,y) = w(x(z),y(z))$ and by the chain rule,

$$ \frac{d w}{dz} = w_x x_z + w_y y_z$$

The next notation uses the fact that $w = w(x,y)$ and so $w_y$ means take partial derivative w.r.t $y$. In this case we have,

$$ \left(\frac{d w}{dz}\right)_y = \frac{\partial}{\partial y}\left(w_x x_z + w_y y_z\right)$$

Now carry out the differentiation again, using a combination of chain rule and product rule. You will get the following,

$$ x_z w_{xy} + y_z w_{yy}$$