Differentiating Inverse Functions - AP Calculus AB Study Guide

The Core Idea: Differentiating Inverse Functions

This topic addresses the relationship between the rate of change of a function and the rate of change of its inverse. If we know the derivative (slope of the tangent line) of a function $f$ at a specific point, we can determine the derivative of its inverse function, $f^{-1}$, at the corresponding point. The fundamental concept is that these derivatives are reciprocals of each other.

Geometrically, the graph of $f^{-1}(x)$ is a reflection of the graph of $f(x)$ across the line $y=x$. This reflection swaps the roles of $x$ and $y$. Consequently, a tangent line with slope $m$ at point $(a,b)$ on the graph of $f$ corresponds to a tangent line with slope $\frac{1}{m}$ at point $(b,a)$ on the graph of $f^{-1}$. This reciprocal relationship is the cornerstone of differentiating inverse functions.

Key Formulas

The rules for differentiating inverse functions are derived from the reciprocal relationship between the slopes of a function and its inverse at corresponding points.

The General Rule for the Derivative of an Inverse

If $f$ is a differentiable function with an inverse function $g(x)=f^{-1}(x)$, and $f^{\prime }(g(x))\ne 0$, then the derivative of the inverse function is given by:

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

Derivatives of Inverse Trigonometric Functions

The following are the specific derivative rules for the six inverse trigonometric functions.

• Inverse Sine:$\frac{d}{dx}(\arcsin (x))=\frac{1}{\sqrt{1-x^2}}$

• Inverse Cosine:$\frac{d}{dx}(\arccos (x))=\frac{-1}{\sqrt{1-x^2}}$

• Inverse Tangent:$\frac{d}{dx}(\arctan (x))=\frac{1}{1+x^2}$

• Inverse Cotangent:$\frac{d}{dx}(\mathrm{arccot}(x))=\frac{-1}{1+x^2}$

• Inverse Secant:$\frac{d}{dx}(\mathrm{arcsec}(x))=\frac{1}{|x|\sqrt{x^2-1}}$

• Inverse Cosecant:$\frac{d}{dx}(\mathrm{arccsc}(x))=\frac{-1}{|x|\sqrt{x^2-1}}$

Understanding the Relationship Between Points

The most critical and often confusing part of applying the general inverse derivative rule is identifying the correct point at which to evaluate the derivative of the original function. The formula $g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$ requires evaluating $f^{\prime }$ not at $x$, but at $g(x)$.

Let's break this down. Suppose we want to find the derivative of the inverse function $g(x)$ at a specific value, say $x=b$. We are looking for $g^{\prime }(b)$.

1. The formula tells us $g^{\prime }(b)=\frac{1}{f^{\prime }(g(b))}$.

2. We need to find the value of $g(b)$. Let's call this value $a$. So, $g(b)=a$.

3. By the definition of an inverse function, if $g(b)=a$, then $f(a)=b$.

4. Substituting $a$ back into our formula, we get $g^{\prime }(b)=\frac{1}{f^{\prime }(a)}$.

This means that to find the derivative of the inverse at $b$, you must first find the input a$ to the original function $f$ that produces the output $b$. Then, you find the derivative of $f$ at that input a` and take its reciprocal.

Core Concepts & Rules

• The derivative of an inverse function $g(x)=f^{-1}(x)$ at a point is the reciprocal of the derivative of the original function $f(x)$ at the corresponding point.

• If a point $(a,b)$ is on the graph of $f(x)$ (meaning $f(a)=b$), then the corresponding point $(b,a)$ is on the graph of $g(x)=f^{-1}(x)$ (meaning $g(b)=a$).

• The derivative of $g$ at $b$ is the reciprocal of the derivative of $f$ at $a$. In symbols: $g^{\prime }(b)=\frac{1}{f^{\prime }(a)}$.

• The general formula is $g^{\prime }(x)=\frac{1}{f^{\prime }(g(x))}$, which requires you to first evaluate the inverse function $g(x)$ before evaluating the derivative $f^{\prime }$.

• The derivatives of the six inverse trigonometric functions ($\arcsin (x)$, $\arccos (x)$, etc.) are specific formulas that must be memorized.

Step-by-Step Example 1: Algebraic Application

Let $f(x)=x^5+3x-2$. If $g(x)=f^{-1}(x)$, find $g^{\prime }(2)$.

Step 1: Identify the corresponding points.

We need to find $g^{\prime }(2)$. Using the formula $g^{\prime }(b)=\frac{1}{f^{\prime }(a)}$, we have $b=2$. We need to find the value $a$ such that $f(a)=b=2$.

$$a^5+3a-2=2$$

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

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

So, we have the point $(1,2)$ on $f(x)$ and the corresponding point $(2,1)$ on $g(x)$. This means $f(1)=2$ and $g(2)=1$.

Step 2: Find the derivative of the original function, $f^{\prime }(x)$.

Using the power rule:

$$f^{\prime }(x)=\frac{d}{dx}(x^5+3x-2)=5x^4+3$$

Step 3: Evaluate $f^{\prime }(x)$ at the correct x-value, $a=1$.

$$f^{\prime }(1)=5(1{)}^4+3=5+3=8$$

Step 4: Take the reciprocal to find $g^{\prime }(2)$.

$$g^{\prime }(2)=\frac{1}{f^{\prime }(1)}=\frac{1}{8}$$

Step-by-Step Example 2: Exam-Style Application (Table Problem)

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

Let $g(x)=f^{-1}(x)$. Find $g^{\prime }(3)$.

Step 1: Identify the corresponding points from the table.

We want to find $g^{\prime }(3)$. This means our $b$ value is 3. We need to find the value $a$ such that $f(a)=3$.

Looking at the table in the $f(x)$ column, we see that $f(4)=3$.

Therefore, $a=4$. This means $g(3)=4$.

Step 2: Find the value of the derivative of $f$ at $a=4$.

We need to find $f^{\prime }(a)=f^{\prime }(4)$.

Looking at the table, we find the entry in the row for $x=4$ and the column for $f^{\prime }(x)$.

The table shows that $f^{\prime }(4)=6$.

Step 3: Take the reciprocal to find $g^{\prime }(3)$.

Using the inverse derivative rule:

$$g^{\prime }(3)=\frac{1}{f^{\prime }(g(3))}=\frac{1}{f^{\prime }(4)}$$

$$g^{\prime }(3)=\frac{1}{6}$$

Using Your Calculator

This topic is primarily tested analytically, meaning you will solve problems by hand. A calculator is not used to directly find the derivative of a general inverse function. However, it can be a powerful tool for checking the components of your work.

For a problem like Example 1 ($f(x)=x^5+3x-2$, find $g^{\prime }(2)$), you can use the calculator to verify the derivative of $f(x)$ at the point you found.

To check $f^{\prime }(1)$:

1. Enter Y_1 = X^5 + 3X - 2.

2. From the main screen, use the numerical derivative function (e.g., nDeriv on a TI-84 or $MATH8$).

3. The syntax is nDeriv(function, variable, value). Enter nDeriv(Y_1, X, 1).

4. The calculator will return $8$, confirming your hand-calculated value for $f^{\prime }(1)$. You would then take the reciprocal to get the final answer, $1/8$.

This method does not find the answer for you, but it helps confirm that the most important calculation in the process is correct.

AP Exam Quick Hit

Common Question Types

• Given a Function, Find the Inverse Derivative at a Point: You are given an explicit function, $f(x)$, and asked to find the derivative of its inverse, $g(x)$, at a specific point $x=b$. This requires you to first solve $f(a)=b$ for $a$, then find $f^{\prime }(a)$, and finally take the reciprocal. (See Example 1).

• Given a Table, Find the Inverse Derivative at a Point: You are given a table of values for $f(x)$ and $f^{\prime }(x)$ and asked to find $g^{\prime }(b)$. This tests your understanding of the concept without complex algebra, as you must look up the correct values in the table. (See Example 2).

• Derivative of an Inverse Trig Function (with Chain Rule): You are asked to find the derivative of a function that is a composition of an inverse trigonometric function and another function, such as $h(x)=\arctan (e^x)$ or $k(x)=\arcsin (x^2)$. This requires knowing the specific inverse trig derivative rule and correctly applying the chain rule.

Common Mistakes

• Plugging in the Wrong Value: The most frequent error is calculating $\frac{1}{f^{\prime }(b)}$ instead of the correct $\frac{1}{f^{\prime }(a)}$, where $f(a)=b$. For instance, in Example 1, a common mistake is to calculate $f^{\prime }(2)$ instead of $f^{\prime }(1)$.

• Forgetting the Reciprocal: Students correctly find the value of $f^{\prime }(a)$ but then forget the final step of taking the reciprocal. They might report the answer as $8$ instead of $1/8$ in Example 1.

• Inverse Trig Sign Errors: Confusing the derivatives of $\arccos (x)$, $\mathrm{arccot}(x)$, and $\mathrm{arccsc}(x)$ with their positive cofunction counterparts. Remember that all the "co-" inverse functions have negative derivatives.

• Omitting the Chain Rule: When differentiating a composite function like $\arctan (u)$, students often find $\frac{1}{1+u^2}$ but forget to multiply by $u^{\prime }=\frac{du}{dx}$.