In one line
If \(z=f(x,y)\) with \(x=x(t)\) and \(y=y(t)\), then \(\dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\dfrac{dy}{dt}\): add the contributions of each path.
Explained simply
Imagine your phone bill \(z\) depends on data used \(x\) and minutes called \(y\), and both grow with time \(t\). The total rate at which your bill grows is: (cost per GB × GB per day) + (cost per minute × minutes per day). Each path from \(t\) to \(z\) contributes a product, and you add the paths. This is the chain rule for several variables.
Technically, if \(z=f(x,y)\) has continuous partial derivatives and \(x=\phi(t)\), \(y=\psi(t)\) are differentiable, then \(z\) is a function of \(t\) alone (a composite function) and \(\dfrac{dz}{dt}=z_x\dfrac{dx}{dt}+z_y\dfrac{dy}{dt}\). The derivative on the left uses ordinary \(d\) because \(z\) finally depends on one variable \(t\).
How it works
- Draw the dependency tree: \(z\to x,y\to t\).
- Find \(z_x\), \(z_y\) (partial) and \(\dfrac{dx}{dt},\dfrac{dy}{dt}\) (ordinary).
- Multiply along each branch and add.
- Substitute \(x,y\) in terms of \(t\) if a final answer in \(t\) is required.
- For three intermediate variables \(x,y,z\to t\), add a third term \(u_z\dfrac{dz}{dt}\).
Key formula
\(\dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\dfrac{dy}{dt}\); for \(u=f(x,y,z)\): \(\dfrac{du}{dt}=u_x\dfrac{dx}{dt}+u_y\dfrac{dy}{dt}+u_z\dfrac{dz}{dt}\).
Quick example
Let \(z=x^2y\), \(x=t^2\), \(y=t^3\).
\(z_x=2xy\), \(z_y=x^2\), \(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2\).
\(\dfrac{dz}{dt}=2xy(2t)+x^2(3t^2)=2(t^2)(t^3)(2t)+(t^4)(3t^2)=4t^6+3t^6=7t^6\).
Check directly: \(z=t^4\cdot t^3=t^7\), so \(\dfrac{dz}{dt}=7t^6\). Correct.
Real use: if the radius of a cylinder grows at 0.1 cm/s and its height shrinks at 0.2 cm/s, the rate of change of volume \(V=\pi r^2h\) is \(2\pi rh(0.1)+\pi r^2(-0.2)\) cm³/s.
Common mistakes
- Writing \(\dfrac{\partial z}{\partial t}\) when \(z\) depends on \(t\) only; use \(\dfrac{dz}{dt}\).
- Missing one branch of the tree, especially when there are three intermediate variables.
- Substituting \(x(t)\), \(y(t)\) into \(z\) before finding \(z_x\), \(z_y\), which mixes the two methods.
How MU asks it
3–5-mark parts: "If \(u=x^2+y^2+z^2\), \(x=e^t\), \(y=e^t\sin t\), \(z=e^t\cos t\), find \(\dfrac{du}{dt}\)" (answer \(4e^{2t}\)), sometimes "verify by direct substitution". For full marks write the chain-rule formula first, list each derivative, substitute, and simplify; if asked to verify, show both methods give the same answer.