Recipe

Apply the multivariable chain rule

Use when: z=f(x,y) and each input is itself a function of one or more other variables, and you need the rate of change of z with respect to one of those underlying variables

Goal: a correct sum of partial-derivative products, one term for every path from z down to the variable being differentiated

Taught in The multivariable chain rule

1. 1

   Put z at the top, its immediate inputs (like x, y) one level below, and each of those inputs' own inputs (like t) another level below that. Draw a branch for every arrow.

   Example

   For z=x^2+y^2 with x=x(t), y=y(t): z has two branches down to x and y, and each of those has one branch down to t. That gives two full paths from z to t: z\to x\to t and z\to y\to t.

   Why: the chain rule sums one term per PATH through this tree, so drawing it first prevents missing or double-counting a path.

2. 2

   Look at the tree from the last step to see whether one variable or two sits at the bottom.

   • only one variable, like t, sits at the bottom — find the single ordinary derivative dz/dt; every branch feeds into that one sum.
   • two or more variables, like s and t, sit at the bottom — find a SEPARATE partial derivative for each one, holding the others fixed while building its own sum.

   Why: which derivative you are computing determines which bottom-level variable each branch's derivative is taken with respect to.

3. 3

   For every path from z to the target variable, multiply together the derivative on each edge of that path. Then sum all the paths' products.

   Example

   Continuing z=x^2+y^2, x=e^t\cos t, y=e^t\sin t: z_x=2x, z_y=2y, \dfrac{dx}{dt}=e^t\cos t-e^t\sin t, \dfrac{dy}{dt}=e^t\sin t+e^t\cos t. Multiply along each branch and add: \frac{dz}{dt}=z_x\frac{dx}{dt}+z_y\frac{dy}{dt}

   Why: each branch chains how much a nudge in the bottom variable moves the middle variable, times how much that middle variable moves z. Adding combines every route z has of feeling the bottom variable move.

4. 4

   Plug the specific value of t into every piece from the previous step, including x and y themselves wherever they still appear, then combine.

   Example

   At t=0: x=e^0\cos0=1, y=e^0\sin0=0, dx/dt=1-0=1, dy/dt=0+1=1. So \dfrac{dz}{dt}\Big|_{t=0}=2(1)(1)+2(0)(1)=2.

   Why: the chain rule's formula is a general pattern in symbols. A specific numeric answer requires plugging in the specific point where the rate of change is actually being asked for.