The Chain Rule: Fresh Take

  • Use the chain rule when you have one independent variable
  • Apply the chain rule with multiple changing variables
  • Find derivatives implicitly when working with functions of multiple variables

Chain Rules for One or Two Independent Variables

The Main Idea 

The chain rule extends to multivariable functions, but now we need different versions depending on how many independent variables we have.

Chain Rule for One Independent Variable

If [latex]z = f(x, y)[/latex] where [latex]x = g(t)[/latex] and [latex]y = h(t)[/latex]:

[latex]\dfrac{dz}{dt}=\dfrac{\partial z}{\partial x}\cdot\dfrac{dx}{dt}+\dfrac{\partial z}{\partial y}\cdot\dfrac{dy}{dt}[/latex]

Notice: We have two terms because [latex]z[/latex] depends on two intermediate variables ([latex]x[/latex] and [latex]y[/latex]). Each term represents one path from [latex]z[/latex] to [latex]t[/latex].

Chain Rule for Two Independent Variables

If [latex]z = f(x, y)[/latex] where [latex]x = g(u, v)[/latex] and [latex]y = h(u, v)[/latex]:

[latex]\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}[/latex]

[latex]\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}[/latex]

Key Difference

  • One independent variable: Use ordinary derivative [latex]\frac{dz}{dt}[/latex]
  • Two independent variables: Use partial derivatives [latex]\frac{\partial z}{\partial u}[/latex] and [latex]\frac{\partial z}{\partial v}[/latex]

Tree Diagrams

Draw branches showing all paths from [latex]z[/latex] to the final variable. Each complete path gives one term. Add all terms together.

Problem-Solving Strategy

Calculate all needed partial derivatives. Apply the chain rule formula. Substitute the intermediate variable definitions. Simplify the result.

Calculate [latex]dz/dt[/latex] given the following functions. Express the final answer in terms of [latex]t[/latex].

[latex]z=f(x, y)=x^{2}-3xy+2y^{2}, x=x(t)=3\sin 2t, y=y(t)=4\cos 2t[/latex]

Calculate [latex]\partial x/\partial u[/latex] and [latex]\partial z/\partial v[/latex] given the following functions:

[latex]z=f(x,y)=\frac{2x-y}{x+3y},x(u,v)=e^{2u}\cos 3v, y(u,v)=e^{2u}\sin 3v[/latex]

Watch the following video to see the worked solution to the above example.

The Generalized Chain Rule

The Main Idea 

The chain rule extends to any number of variables. The pattern stays consistent: add one term for each intermediate variable.

The General Formula

Let [latex]w = f(x_1, x_2, \ldots, x_m)[/latex] depend on [latex]m[/latex] variables, where each [latex]x_i[/latex] depends on [latex]n[/latex] variables [latex]t_1, t_2, \ldots, t_n[/latex]:

[latex]\dfrac{\partial w}{\partial t_j}=\dfrac{\partial w}{\partial x_1}\dfrac{\partial x_1}{\partial t_j}+\dfrac{\partial w}{\partial x_2}\dfrac{\partial x_2}{\partial t_j}+\cdots+\dfrac{\partial w}{\partial x_m}\dfrac{\partial x_m}{\partial t_j}[/latex]

The number of terms equals the number of intermediate variables. If [latex]w[/latex] depends on three intermediate variables ([latex]x[/latex], [latex]y[/latex], [latex]z[/latex]), you get three terms.

Tree Diagrams Help

Draw branches from [latex]w[/latex] to each intermediate variable ([latex]x[/latex], [latex]y[/latex], [latex]z[/latex]). From each intermediate variable, draw branches to each final variable ([latex]u[/latex], [latex]v[/latex]). Each complete path gives one term in the formula.

Problem-Solving Strategy

Identify all intermediate variables. Calculate all needed partial derivatives. Apply the chain rule formula—one term per intermediate variable. Substitute and simplify.

Calculate [latex]\partial w/\partial u[/latex] and [latex]\partial w/\partial v[/latex] given the following functions:[latex]\hspace{8cm} \begin{align}w&=f(x,y,z)=\frac{x+2y-4z}{2x-y+3z} \\ x&=x(u,v) = e^{2u}\cos 3v \\ y&=y(u,v) = e^{2u}\sin 3v \\ z&=z(u,v) = e^{2u}. \end{align}[/latex]

Watch the following video to see the worked solution to the above example.

Create a tree diagram for the case when

[latex]\large{w=f(x,y),x=x(t,u,v),y=y(t,u,v)}[/latex]

and write out the formulas for the three partial derivatives of [latex]w[/latex].

Implicit Differentiation

The Main Idea 

Partial derivatives give us a shortcut for implicit differentiation. Instead of differentiating the entire equation and solving, just calculate two partial derivatives and plug them into a formula.

For Two Variables

If [latex]f(x, y) = 0[/latex] defines [latex]y[/latex] implicitly as a function of [latex]x[/latex]:

[latex]\dfrac{dy}{dx}=-\dfrac{\partial f/\partial x}{\partial f/\partial y}[/latex]

(provided [latex]f_y \neq 0[/latex])

For Three Variables

If [latex]f(x, y, z) = 0[/latex] defines [latex]z[/latex] implicitly as a function of [latex]x[/latex] and [latex]y[/latex]:

[latex]\dfrac{\partial z}{\partial x}=-\dfrac{\partial f/\partial x}{\partial f/\partial z}[/latex] and [latex]\dfrac{\partial z}{\partial y}=-\dfrac{\partial f/\partial y}{\partial f/\partial z}[/latex]

(provided [latex]f_z \neq 0[/latex])

Why This Works

The formula comes from the chain rule. When you differentiate [latex]f(x, y) = 0[/latex] with respect to [latex]x[/latex], you get [latex]f_x + f_y \frac{dy}{dx} = 0[/latex]. Solving for [latex]\frac{dy}{dx}[/latex] gives the formula.

Problem-Solving Strategy

Define [latex]f[/latex] using the left side of the equation. Calculate the needed partial derivatives. Plug into the formula. Simplify.

Find [latex]dy/dx[/latex] if [latex]y[/latex] is defined implicitly as a function of [latex]x[/latex] by the equation [latex]x^{2}+xy-y^{2}+7x-3y-26=0[/latex]. What is the equation of the tangent line to the graph of this curve at point [latex](3, -2)[/latex]?

Watch the following video to see the worked solution to the above example.