In one line
If \(z=f(x,y)\) with \(x=x(u,v)\), \(y=y(u,v)\), then \(z_u=z_xx_u+z_yy_u\) and \(z_v=z_xx_v+z_yy_v\); every path from \(z\) to the new variable contributes one product.
Explained simply
This is a change of coordinates. A map can label a place by (east, north) or by (distance, direction). If a quantity is known in one set of coordinates, the chain rule tells you its rates of change in the other set. Engineers do this all the time, for example switching from \(x,y\) to polar coordinates \(r,\theta\) for circular plates or pipes.
Technically, if \(z=f(x,y)\) and \(x=g(u,v)\), \(y=h(u,v)\) all have continuous partial derivatives, then \(\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}\), and similarly for \(v\). All derivatives here are partial because there are always two independent variables.
How it works
- Draw the tree: \(z\to(x,y)\to(u,v)\).
- For each new variable, go along every branch, multiply, and add.
- If the function is given as \(u=f(r,s,t)\) with \(r,s,t\) defined in terms of \(x,y,z\), introduce those letters first and compute \(r_x,s_x,t_x\), etc.
- Combine the results as the question asks (sum, squares, etc.).
Key formula
\(z_u=z_xx_u+z_yy_u,\quad z_v=z_xx_v+z_yy_v\).
Polar case: \(x=r\cos\theta,\ y=r\sin\theta\) gives \(z_r=z_x\cos\theta+z_y\sin\theta\), \(z_\theta=-z_xr\sin\theta+z_yr\cos\theta\), and hence \(z_x^2+z_y^2=z_r^2+\dfrac1{r^2}z_\theta^2\).
Quick example
If \(u=f(x-y,\,y-z,\,z-x)\), prove \(u_x+u_y+u_z=0\).
Let \(r=x-y\), \(s=y-z\), \(t=z-x\), so \(u=f(r,s,t)\).
- \(u_x=u_r(1)+u_s(0)+u_t(-1)=u_r-u_t\)
- \(u_y=u_r(-1)+u_s(1)+u_t(0)=-u_r+u_s\)
- \(u_z=u_r(0)+u_s(-1)+u_t(1)=-u_s+u_t\)
Adding: \(u_x+u_y+u_z=0\). Proved. Notice we never needed to know what \(f\) is.
Common mistakes
- Forgetting zero terms and then losing track; write all three products even when one is zero.
- Mixing up \(x_u\) and \(u_x\) in the tree.
- In the polar result, forgetting the \(\dfrac1{r^2}\) factor with \(z_\theta^2\).
How MU asks it
Typical 5/6-mark parts: "If \(u=f\left(\dfrac{x}{y},\dfrac{y}{z},\dfrac{z}{x}\right)\), prove \(xu_x+yu_y+zu_z=0\)", "If \(z=f(x,y)\), \(x=r\cos\theta\), \(y=r\sin\theta\), prove \(z_x^2+z_y^2=z_r^2+\dfrac{1}{r^2}z_\theta^2\)", or "If \(z=f(x,y)\), \(x=e^u+e^{-v}\), \(y=e^{-u}-e^v\), prove \(z_u-z_v=xz_x-yz_y\)". Define the new letters, write each chain-rule line in full, then combine.