The Chain Rule - AP Calculus AB Study Guide

The Core Idea: The Chain Rule

The fundamental rules of differentiation allow us to find derivatives of simple functions like polynomials, trigonometric functions, and exponentials. However, many functions in mathematics are composite functions, where one function is nested inside another. For example, in the function $h(x)=\sin (x^2)$, the function $g(x)=x^2$ is the "input" for the function $f(u)=\sin (u)$. We cannot simply find the derivative of the "outside" part and ignore the "inside" part.

The Chain Rule provides a necessary and powerful method for finding the derivative of such composite functions. It formalizes the process of accounting for the rate of change of the inner function and how that impacts the rate of change of the outer function. It is the tool that "unpacks" the layers of a composite function to find its overall rate of change.

Key Formulas

The Chain Rule can be expressed in two common and equivalent ways, based on the provided Essential Knowledge.

Function Notation

This form is most useful when working with functions written in the form $h(x)=f(g(x))$, where $f$ is the "outer" function and $g$ is the "inner" function.

• Formula: The derivative of $h(x)=f(g(x))$ is given by:

  $$h^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$$

• Explanation: This formula states that you first take the derivative of the outer function $f$, leaving the inner function $g(x)$ unchanged inside it. Then, you multiply this entire result by the derivative of the inner function, $g^{\prime }(x)$.

Leibniz Notation

This form is useful for seeing the "chain" of rates of change, especially when variables are defined in terms of each other.

• Formula: If $y$ is a function of $u$, written as $y=f(u)$, and $u$ is a function of $x$, written as $u=g(x)$, then the derivative of $y$ with respect to $x$ is:

  $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$

• Explanation: This notation highlights how the rate of change of $y$ with respect to $x$ ($\frac{dy}{dx}$) is the product of the rate of change of $y$ with respect to its immediate input `u$ ($\frac{dy}{du}$) and the rate of change of $u$ with respect to $x$ ($\frac{du}{dx}$).

Understanding Composition

The most critical skill for applying the Chain Rule is correctly identifying the "outer" and "inner" functions in a composition. A composite function $h(x)=f(g(x))$ is formed when the output of one function, $g(x)$, becomes the input of another function, $f(x)$.

• Identifying Layers: To find the derivative of $h(x)=\sqrt{x^2+1}$, you must first recognize its structure.

  • The "last operation" performed is the square root. This is the outer function: $f(u)=\sqrt{u}=u^{1/2}$.

  • The expression inside the square root is the inner function: $g(x)=x^2+1$.

• Multiple Compositions: The Chain Rule can be extended to compositions of more than two functions. For a function like $k(x)={\cos }^3(4x)$, which is equivalent to $k(x)=(\cos (4x){)}^3$, there are three layers:

  • Outermost function:$f(u)=u^3$

  • Middle function:$g(v)=\cos (v)$

  • Innermost function:$h(x)=4x$

  The derivative would be an extension of the rule: $k^{\prime }(x)=f^{\prime }(g(h(x)))\cdot g^{\prime }(h(x))\cdot h^{\prime }(x)$.

Core Concepts & Rules

• The Chain Rule is the specific method required to differentiate composite functions of the form $f(g(x))$.

• The derivative of a composite function is the product of two parts: the derivative of the outer function (evaluated at the original inner function) and the derivative of the inner function.

• The Chain Rule can be applied repeatedly for functions with multiple layers of composition (e.g., $f(g(h(x)))$).

• The Chain Rule is not a standalone rule; it must be combined with all other differentiation rules (Product, Quotient, Power, etc.) when a composite function appears as part of a larger expression. For example, to differentiate $x^2\tan (5x)$, you would need the Product Rule first, and then the Chain Rule to find the derivative of $\tan (5x)$.

Step-by-Step Example 1: Basic Application

Problem: Find the derivative of $h(x)=(3x^4-7{)}^5$.

Step 1: Identify the outer and inner functions.

This function has the form $f(g(x))$.

• The outer function is the power of 5: $f(u)=u^5$.

• The inner function is the polynomial inside the parentheses: $g(x)=3x^4-7$.

Step 2: Find the derivatives of the outer and inner functions.

• Derivative of the outer function: $f^{\prime }(u)=5u^4$.

• Derivative of the inner function: $g^{\prime }(x)=12x^3$.

Step 3: Apply the Chain Rule formula: $h^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$.

• First, find $f^{\prime }(g(x))$. This means we take the derivative of the outer function, $5u^4$, and substitute the original inner function $g(x)$ back in for $u$.

  $f^{\prime }(g(x))=5(3x^4-7{)}^4$

• Next, multiply by the derivative of the inner function, $g^{\prime }(x)$.

  $h^{\prime }(x)=5(3x^4-7{)}^4\cdot (12x^3)$

Step 4: Simplify the result.

• Combine the terms outside the parentheses.

  $h^{\prime }(x)=60x^3(3x^4-7{)}^4$

Step-by-Step Example 2: Exam-Style Application

Problem: The functions $f$ and $g$ are differentiable. The table below gives values of the functions and their derivatives at selected values of $x$.

If $h(x)=f(g(x))$, what is the value of $h^{\prime }(1)$?

Step 1: Write down the general formula for the derivative of $h(x)$.

• Using the Chain Rule, the derivative of $h(x)=f(g(x))$ is:

  $h^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$

Step 2: Evaluate the derivative formula at the specific point, $x=1$.

• Substitute $x=1$ into the derivative formula:

  $h^{\prime }(1)=f^{\prime }(g(1))\cdot g^{\prime }(1)$

Step 3: Use the table to find the values needed, working from the inside out.

• First, we need the value of the inner function, $g(1)$. Look at the table in the row for $x=1$ and the column for $g(x)$.

  $g(1)=3$

Step 4: Substitute this value back into the expression for $h^{\prime }(1)$.

• Replace $g(1)$ with its value, 3:

  $h^{\prime }(1)=f^{\prime }(3)\cdot g^{\prime }(1)$

Step 5: Use the table to find the remaining derivative values.

• We need $f^{\prime }(3)$. Look at the row for $x=3$ and the column for $f^{\prime }(x)$.

  $f^{\prime }(3)=-6$

• We need $g^{\prime }(1)$. Look at the row for $x=1$ and the column for $g^{\prime }(x)$.

  $g^{\prime }(1)=-2$

Step 6: Calculate the final result.

• Substitute these values into the expression:

  $h^{\prime }(1)=(-6)\cdot (-2)=12$

Using Your Calculator

The Chain Rule is an analytical process that must be performed by hand. A calculator cannot apply the rule for you to find a general derivative function. However, it is an excellent tool for checking your work by calculating the numerical value of a derivative at a specific point.

To verify the derivative of $h(x)=(3x^4-7{)}^5$ at $x=1$:

1. By Hand:

   • Our derivative is $h^{\prime }(x)=60x^3(3x^4-7{)}^4$.

   • Evaluate at $x=1$: $h^{\prime }(1)=60(1{)}^3(3(1{)}^4-7{)}^4=60(3-7{)}^4=60(-4{)}^4=60(256)=15360$.

2. On the Calculator (TI-84 Style):

   • Use the numerical derivative function, often found under the MATH menu (nDeriv) or by pressing $alpha$$window$.

   • The syntax is nDeriv(function, variable, point).

   • Enter the following on your home screen: nDeriv((3X^4 - 7)^5, X, 1)

   • The calculator will return $15360$, confirming that our analytical derivative is correct at that point.

AP Exam Quick Hit

Common Question Types

• Combining with Product/Quotient Rules: You will be asked to find the derivative of a function like $k(x)=x^3\cos (2x)$ or $p(x)=\frac{e^{5x}}{(x^2+1)}$. These require you to apply the Product or Quotient Rule as the main structure, and then use the Chain Rule to differentiate the composite part (e.g., $\cos (2x)$ or $e^{5x}$).

• Table Problems: As shown in Example 2, you will be given a table of function and derivative values and asked to compute the derivative of a composition like $f(g(x))$ or $g(f(x))$ at a specific point. This tests your direct knowledge of the formula $h^{\prime }(a)=f^{\prime }(g(a))\cdot g^{\prime }(a)$.

• Implicit Differentiation: The Chain Rule is the foundation of implicit differentiation. When differentiating an equation with respect to $x$, any term involving $y$ is treated as a composite function, requiring the Chain Rule. For example, the derivative of $y^2$ with respect to $x$ is $2y\cdot \frac{dy}{dx}$.

Common Mistakes

• Forgetting to Multiply by the Inner Derivative: This is the most frequent error. For example, stating that the derivative of $e^{3x}$ is $e^{3x}$ instead of the correct $e^{3x}\cdot 3$.

• Misapplying the Power Rule to Functions: A common mistake is seeing ${\sin }^2(x)$ and writing the derivative as $2\cos (x)$. You must recognize that ${\sin }^2(x)$ is $(\sin x{)}^2$, an "inside" function ($\sin x$) to the power of 2. The correct derivative is $2(\sin x{)}^1\cdot \cos (x)$.

• Confusing Values in Table Problems: In a table problem asking for $f^{\prime }(g(a))\cdot g^{\prime }(a)$, students often mistakenly calculate $f^{\prime }(a)\cdot g^{\prime }(a)$ or $f^{\prime }(g^{\prime }(a))$. It is crucial to work from the inside out: first find the value of $g(a)$, then use that value as the input for $f^{\prime }$.

• Incorrect Derivative of the Outer Function: Students sometimes differentiate the outer function and simultaneously change the inner function. For $h(x)=\sin (x^2)$, the term $f^{\prime }(g(x))$ is $\cos (x^2)$, NOT $\cos (2x)$. The inner function must remain untouched in the first part of the rule.