- Explain and use the chain rule
- Use the chain rule along with other rules to differentiate functions involving powers, products, quotients, and trigonometry
- Use the chain rule to find derivatives when multiple functions are nested together
Deriving the Chain Rule
The process of differentiating basic functions like [latex]x^n, \, \sin x,[/latex], and [latex]\cos x[/latex] as well as products and quotients, has already been established. However, these techniques alone do not address differentiating compositions of functions, such as [latex]h(x)= \sin (x^3)[/latex] or [latex]k(x)=\sqrt{3x^2+1}[/latex]. To derive these, we utilize the chain rule.
The chain rule simplifies the process of differentiating composite functions by breaking down a function into simpler parts. Rather than differentiating the entire composite directly, the chain rule allows us to differentiate each part independently. Specifically, for a composite function [latex]h(x)=f(g(x))[/latex], the derivative is given by the derivative of the outer function [latex]f[/latex] evaluated at the inner function [latex]g(x)[/latex], multiplied by the derivative of the inner function:
[latex]h^{\prime} (x)=f^{\prime} (g(x)) \cdot g^{\prime} (x)[/latex]
To contextualize this, consider [latex]h(x)= \sin (x^3)[/latex].
First, identify the inner function [latex]g(x)=x^3[/latex] and the outer function[latex]f(u)= \sin u[/latex], where [latex]u=g(x)[/latex]. Applying the chain rule, which states the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function, we proceed as follows:
- Differentiate the inner function [latex]g(x)[/latex]:
[latex]g^{\prime} (x) = 3x^2[/latex]
- Differentiate the outer function [latex]f(u)[/latex] with respect to [latex]u[/latex]:
[latex]f^{\prime}(u) = \cos u[/latex]
- Since [latex]u=g(x)=x^3[/latex], substituting that back in, we get:
[latex]f^{\prime} (u) = \cos{(x^3)}[/latex]
- Apply the chain rule:
[latex]h^{\prime} (x)=f^{\prime} (g(x)) \cdot g^{\prime} (x)= \cos{(x^3)} \cdot 3x^2[/latex]
Therefore, the derivative of [latex]h(x)= \sin (x^3)[/latex] is:
Now that we’ve illustrated how to apply the chain rule with a specific example, let’s explore the general formula of the chain rule and see how it applies to various types of composite functions. An informal proof of this concept will follow at the end of this section.
the chain rule
Let [latex]f[/latex] and [latex]g[/latex] be functions. For all [latex]x[/latex] in the domain of [latex]g[/latex] for which [latex]g[/latex] is differentiable at [latex]x[/latex] and [latex]f[/latex] is differentiable at [latex]g(x)[/latex], the derivative of the composite function [latex]h(x)=(f\circ g)(x)=f(g(x))[/latex] is given by:
Alternatively, if [latex]y[/latex] is a function of [latex]u[/latex], and [latex]u[/latex] is a function of [latex]x[/latex], then
For more information, check out this interactive on The Intuitive Notion of the Chain Rule.
How to: Apply the Chain Rule
- Identify the Functions: Begin by identifying the inner function [latex]g(x)[/latex] and the outer function [latex]f(u)[/latex], where [latex]u=g(x)[/latex].
- Derivative of the Outer Function: Compute the derivative of the outer function, [latex]f^{\prime}(u)[/latex], and evaluate it at [latex]g(x)[/latex] to obtain [latex]f^{\prime}(g(x))[/latex].
- Derivative of the Inner Function: Calculate the derivative of the inner function [latex]g^{\prime}(x)[/latex].
- Apply the Chain Rule: Combine these results to find [latex]h^{\prime}(x)[/latex] as follows:
[latex]h^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)[/latex]
When applying the chain rule to compositions involving multiple functions, remember that each function contributes to the derivative. Start from the innermost function and work outward, applying the chain rule iteratively. The derivative of a composite involving three functions, for example, involves taking derivatives step by step, moving from the innermost to the outermost function. Importantly, derivatives are not evaluated at derivatives; they are evaluated at functions.