Chain Rule - Intuition

Given 𝒽(𝑥) = 𝑓(𝑔(𝑥)), then

Indent

Example

Suppose we want to compute the derivative of a composite function 𝒽(𝑥)

𝒽(𝑥) = (5 - 6𝑥)⁵

• 𝑔(𝑥) = 5 - 6𝑥
• 𝑓(𝑥) = 𝑥⁵
• 𝑔’(𝑥) = -6
• 𝑓’(𝑥) = 5𝑥⁴

Apply chain rule

• 𝑑/𝑑𝑥[𝒽(𝑥)] = 𝑑/𝑑𝑥[𝑓(𝑔(𝑥))]
• 𝑑/𝑑𝑥[𝒽(𝑥)] = 𝑓’(𝑔(𝑥)) * 𝑔’(𝑥)
• 𝑑/𝑑𝑥[𝒽(𝑥)] = 5(5 - 6𝑥)⁴ * -6
• 𝑑/𝑑𝑥[𝒽(𝑥)] = -30(5 - 6𝑥)⁴

Multivariable Chain Rule

• $\frac{d}{dt}f(\overrightarrow{v}(t))=\nabla f(\overrightarrow{v}(t))\cdot \overrightarrow{v^{\prime }}(t)$

where:

• 𝑓 is a scalar-valued function
• 𝛻𝑓 is the gradient vector of 𝑓
• 𝑣̅(𝑡) is a vector-valued function (can be a vector of size 1)
• · is the dot product

This is similar to the definition of a directional derivative

Examples

Scalar-Valued Function of 1 Variable

Given a scalar-valued function 𝑓 that takes one variable 𝑥 where it is dependent on 𝑡:

• $\frac{d}{dt}f(x(t))=\left[\begin{array}{c}\frac{𝛿f}{𝛿x}\end{array}\right]\cdot \left[\begin{array}{c}\frac{𝛿x}{𝛿t}\end{array}\right]$
• $\frac{d}{dt}f(x(t))=\frac{𝛿f}{𝛿x}\frac{𝛿x}{𝛿t}$

Scalar-Valued Function of 2 Variables

Given a scalar-valued function 𝑓 that takes two variables 𝑥 and 𝑦 where they are dependent on 𝑡:

• $\frac{d}{dt}f(x(t),y(t))=\left[\begin{array}{c}\frac{𝛿f}{𝛿x}\\ \frac{𝛿f}{𝛿y}\end{array}\right]\cdot \left[\begin{array}{c}\frac{𝛿x}{𝛿t}\\ \frac{𝛿y}{𝛿t}\end{array}\right]$
• $\frac{d}{dt}f(x(t),y(t))=\frac{𝛿f}{𝛿x}\frac{𝛿x}{𝛿t}+\frac{𝛿f}{𝛿y}\frac{𝛿y}{𝛿t}$

multivariable-chain-rule-example.drawio