Differentiating Inverse Functions - AP Calculus BC Study Guide

The Core Idea: Differentiating Inverse Functions

This topic addresses how to find the rate of change (the derivative) of a function's inverse without needing to find an explicit equation for the inverse function itself. The fundamental concept is that a deep relationship exists between the derivative of a function, $f$, and the derivative of its inverse, $g$. This relationship is not one of inversion, but rather one of reciprocation.

The derivative of the inverse function at a specific point is the reciprocal of the derivative of the original function, evaluated at the corresponding point. This allows us to calculate the slope of the tangent line to the inverse function's graph by using information readily available from the original function, $f$, and its derivative, $f^{\prime }$. This technique is powerful because finding an algebraic expression for an inverse function can often be difficult or impossible.

Key Formulas

The primary formula for differentiating an inverse function is derived directly from the relationship between a function and its inverse. If the function $g$ is the inverse of the function $f$, then the derivative of $g$ is given by:

$$g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$$

Where:

• $g^{\prime }(x)$ is the derivative of the inverse function.

• $f^{\prime }$ is the derivative of the original function.

• $g(x)$ is the inverse function. The value of $g(x)$ serves as the input for the derivative of the original function, $f^{\prime }$.

Understanding the Relationship

The most critical nuance in applying the formula for the derivative of an inverse function is understanding the input for $f^{\prime }$. The formula requires evaluating $f^{\prime }$ not at $x$, but at $g(x)$. This means we are connecting points between the two functions.

Suppose we have a point $(a,b)$ on the graph of $f(x)$. This means that $f(a)=b$. By the definition of an inverse function, a corresponding point $(b,a)$ must exist on the graph of $g(x)$, which means $g(b)=a$.

Now, let's find the derivative of the inverse function $g$ at the point $x=b$. Using the formula:

$$g^{\prime }(b)=\frac{1}{f^{\prime }(g(b))}$$

Since we established that $g(b)=a$, we can substitute this into the formula:

$$g^{\prime }(b)=\frac{1}{f^{\prime }(a)}$$

This result reveals the core concept: the slope of the tangent line to the inverse function $g$ at $x=b$ is the reciprocal of the slope of the tangent line to the original function $f$ at $x=a$. The key is to correctly identify the corresponding points $(a,b)$ on $f$ and $(b,a)$ on $g$.

Core Concepts & Rules

• The derivative of an inverse function is directly related to the derivative of the original function.

• This relationship is defined by reciprocation: the derivative of the inverse is the reciprocal of the original function's derivative, evaluated at a corresponding value.

• The formal rule states that if $g$ is the inverse of $f$, then $g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$.

• To find the derivative of an inverse $g$ at a point $b$, i.e., $g^{\prime }(b)$, you must first find the value $a$ such that $f(a)=b$.

• Once $a$ is found, the derivative is calculated as $g^{\prime }(b)=\frac{1}{f^{\prime }(a)}$. You do not need to find an explicit formula for $g(x)$.

Step-by-Step Example 1: Basic Application

Problem: Let $f(x)=x^3+4x+2$. If $g(x)$ is the inverse function of $f(x)$, what is the value of $g^{\prime }(7)$?

Step 1: State the Goal and the Formula

Our goal is to find $g^{\prime }(7)$. The formula for the derivative of an inverse function is $g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$. For our specific problem, this is:

$$g^{\prime }(7)=\frac{1}{f^{\prime }(g(7))}$$

Step 2: Find the Value of $g(7)$

We do not have an equation for $g(x)$. Instead, we use the relationship between $f$ and $g$. Let $g(7)=a$. By the definition of an inverse function, this means $f(a)=7$. We set up the equation:

$$a^3+4a+2=7$$

$$a^3+4a-5=0$$

By inspection, we can see that $a=1$ is the solution, since $1^3+4(1)-5=1+4-5=0$.

Therefore, $g(7)=1$.

Step 3: Find the Derivative of $f(x)$

We need to calculate $f^{\prime }(x)$ to use in our formula.

$$f(x)=x^3+4x+2$$

$$f^{\prime }(x)=3x^2+4$$

Step 4: Evaluate $f^{\prime }$ at $g(7)$

From Step 2, we know $g(7)=1$. We substitute this value into $f^{\prime }(x)$:

$$f^{\prime }(g(7))=f^{\prime }(1)=3(1{)}^2+4=3+4=7$$

Step 5: Calculate the Final Result

Now we substitute the result from Step 4 back into our formula from Step 1:

$$g^{\prime }(7)=\frac{1}{f^{\prime }(g(7))}=\frac{1}{7}$$

The value of $g^{\prime }(7)$ is $\frac{1}{7}$.

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

Problem: The functions $f$ and $g$ are differentiable, and $g$ is the inverse of $f$. The table below gives values of $f(x)$ and $f^{\prime }(x)$ for selected values of $x$. Find the value of $g^{\prime }(4)$.

Step 1: State the Formula

We need to find $g^{\prime }(4)$. The governing formula is:

$$g^{\prime }(4)=\frac{1}{f^{\prime }(g(4))}$$

Step 2: Find the Value of $g(4)$ using the Table

We need to find the input to $f$ that produces an output of 4. We are looking for a value $a$ such that $f(a)=4$. We scan the $f(x)$ column in the table for the value 4.

The table shows that $f(1)=4$.

By the definition of an inverse function, if $f(1)=4$, then $g(4)=1$.

Step 3: Find the Value of $f^{\prime }(g(4))$ using the Table

From Step 2, we know that $g(4)=1$. Therefore, we need to find $f^{\prime }(1)$.

We look at the table for the row where $x=1$ and find the corresponding value in the $f^{\prime }(x)$ column.

The table shows that $f^{\prime }(1)=-5$.

Step 4: Calculate the Final Result

Substitute the value from Step 3 into the formula from Step 1:

$$g^{\prime }(4)=\frac{1}{f^{\prime }(g(4))}=\frac{1}{f^{\prime }(1)}=\frac{1}{-5}=-\frac{1}{5}$$

The value of $g^{\prime }(4)$ is $-\frac{1}{5}$.

Using Your Calculator

This topic is primarily analytical, and the formula $g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$ must be applied without direct calculator assistance. A calculator is not used to find the derivative of an inverse function directly.

However, a calculator can be a useful tool for two specific intermediate steps:

1. Solving for $g(x)$: In problems like Example 1, you may need to solve an equation like $f(a)=b$. If the equation is not easily solvable by inspection (e.g., $a^5+3a-10=0$), you can use your calculator's graphing or equation-solving features. For example, you could graph $Y_1=X^5+3X-10$ and find the x-intercept (zero) to determine the value of $a$.

2. Verifying Derivatives: You can use the numerical derivative feature of your calculator (e.g., nDeriv or $d/dx$) to check your by-hand calculation of $f^{\prime }(a)$. If you calculated $f^{\prime }(1)=7$ in Example 1, you could use your calculator to compute the numerical derivative of $f(x)=x^3+4x+2$ at $x=1$ to confirm your answer.

The calculator serves as a tool for verification and for handling complex algebra, not for executing the core calculus concept itself.

AP Exam Quick Hit

Common Question Types

• Given an explicit function: You will be given a function, such as $f(x)=e^x+2x$, and asked to find the derivative of its inverse $g$ at a specific point, like $g^{\prime }(1)$. This requires you to solve $f(a)=1$ (in this case, $a=0$) and then compute $1/f^{\prime }(0)$.

• Given a table of values: You will be provided a table with values for $f(x)$ and $f^{\prime }(x)$ and asked to find the derivative of the inverse $g$ at a point. This tests your ability to use the formula by correctly extracting values from the table, as shown in Example 2.

Common Mistakes

• Incorrect Input for $f^{\prime }$: The most frequent error is calculating $\frac{1}{f^{\prime }(b)}$ instead of the correct $\frac{1}{f^{\prime }(a)}$ (where $f(a)=b$). Students forget to first find the corresponding x-value on the original function before evaluating the derivative.

• Attempting to Find the Inverse Function: Students often waste valuable time trying to algebraically find an explicit formula for $g(x)$. For most AP-level problems, this is either impossible or far too time-consuming. The formula is specifically designed to avoid this step.

• Reciprocal of the Function, Not the Derivative: A less common but critical error is confusing $\frac{1}{f^{\prime }(g(x))}$ with $\frac{1}{f(g(x))}$. Remember that the relationship is between the derivatives (slopes), not the function values themselves.

• Notation Confusion: Confusing the inverse function notation $f^{-1}(x)$ with the reciprocal $\frac{1}{f(x)}$. These are entirely different concepts.