Differentiating Inverse Trigonometric Functions - AP Calculus AB Study Guide

The Core Idea: Differentiating Inverse Trigonometric Functions

This topic extends our knowledge of differentiation to a new class of functions: inverse trigonometric functions. While the derivatives of standard trigonometric functions like $\sin (x)$ and $\cos (x)$ result in other trigonometric functions, the derivatives of their inverses, such as $\arcsin (x)$ and $\arctan (x)$, result in algebraic functions. The core task is to learn the specific derivative rules for these inverse functions and, crucially, how to combine them with the chain rule when the input to the function is more complex than just $x$. Mastering this allows us to find the instantaneous rate of change for a broader set of functions that model various phenomena.

Key Formulas

The AP Calculus AB curriculum requires you to know the derivatives of three specific inverse trigonometric functions. These rules should be memorized.

1. Derivative of Inverse Sine (arcsin)

   $$\frac{d}{dx}[\arcsin (x)]=\frac{1}{\sqrt{1-x^2}}$$

2. Derivative of Inverse Cosine (arccos)

   $$\frac{d}{dx}[\arccos (x)]=-\frac{1}{\sqrt{1-x^2}}$$

3. Derivative of Inverse Tangent (arctan)

   $$\frac{d}{dx}[\arctan (x)]=\frac{1}{1+x^2}$$

Understanding The Chain Rule with Inverse Trig Functions

On the AP Exam, it is rare to be asked for the derivative of simply $\arcsin (x)$. Instead, you will almost always need to apply the chain rule, where the argument of the inverse trigonometric function is itself a function, which we can call $u$.

The rule is: find the derivative of the "outside" function (the inverse trig function) with respect to the "inside" function ($u$), and then multiply by the derivative of the "inside" function ($du/dx$).

The general forms are:

1. Chain Rule with Inverse Sine

   If $y=\arcsin (u)$, where $u$ is a function of $x$, then:

   $$\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$$

2. Chain Rule with Inverse Cosine

   If $y=\arccos (u)$, where $u$ is a function of $x$, then:

   $$\frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$$

3. Chain Rule with Inverse Tangent

   If $y=\arctan (u)$, where $u$ is a function of $x$, then:

   $$\frac{dy}{dx}=\frac{1}{1+u^2}\cdot \frac{du}{dx}$$

Core Concepts & Rules

• Algebraic Derivatives: The derivatives of inverse trigonometric functions are algebraic expressions, not trigonometric ones. For example, the derivative of $\arcsin (x)$ involves a square root and a polynomial, not $\cos (x)$.

• Sine/Cosine Relationship: The derivative of $\arccos (u)$ is the negative of the derivative of $\arcsin (u)$. Memorizing one helps you know the other.

• Chain Rule is Key: The most critical skill for this topic is identifying the inner function $u$ and correctly finding its derivative, $du/dx$, to multiply at the end.

• Notation: Remember that $\arcsin (x)$ is the same as ${\sin }^{-1}(x)$. This is inverse function notation, not an exponent. ${\sin }^{-1}(x)\ne \frac{1}{\sin (x)}$.

Step-by-Step Example 1: Basic Application

Problem: Find the derivative of $f(x)=\arctan (4x^2)$.

Step 1: Identify the outer function and the inner function ($u$).

The outer function is $\arctan (\cdot )$.

The inner function is $u=4x^2$.

Step 2: Find the derivative of the inner function, $du/dx$.

Using the power rule:

$$u=4x^2⟹\frac{du}{dx}=8x$$

Step 3: Apply the chain rule formula for $\arctan (u)$.

The formula is $\frac{d}{dx}[\arctan (u)]=\frac{1}{1+u^2}\cdot \frac{du}{dx}$.

Step 4: Substitute $u$ and $du/dx$ into the formula.

Substitute $u=4x^2$ and $\frac{du}{dx}=8x$.

$$f^{\prime }(x)=\frac{1}{1+(4x^2{)}^2}\cdot (8x)$$

Step 5: Simplify the expression.

Be careful with algebra. Square the entire term $(4x^2)$.

$$(4x^2{)}^2=4^2\cdot (x^2{)}^2=16x^4$$

Now, place the $8x$ term in the numerator.

$$f^{\prime }(x)=\frac{8x}{1+16x^4}$$

This is the final derivative.

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

Problem: Find the equation of the line tangent to the graph of $g(x)=x\cdot \arcsin (x)$ at $x=\frac{1}{2}$.

Step 1: Find the derivative, $g^{\prime }(x)$, using the Product Rule.

The function is a product of $x$ and $\arcsin (x)$. Let $f(x)=x$ and $h(x)=\arcsin (x)$.

The Product Rule is $g^{\prime }(x)=f^{\prime }(x)h(x)+f(x)h^{\prime }(x)$.

• $f^{\prime }(x)=1$

• $h^{\prime }(x)=\frac{d}{dx}[\arcsin (x)]=\frac{1}{\sqrt{1-x^2}}$

Now, apply the rule:

$$g^{\prime }(x)=(1)\cdot \arcsin (x)+x\cdot \left(\frac{1}{\sqrt{1-x^2}}\right)$$

$$g^{\prime }(x)=\arcsin (x)+\frac{x}{\sqrt{1-x^2}}$$

Step 2: Evaluate the derivative at $x=\frac{1}{2}$ to find the slope ($m$) of the tangent line.

$$m=g^{\prime }\left(\frac{1}{2}\right)=\arcsin \left(\frac{1}{2}\right)+\frac{\frac{1}{2}}{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}$$

Recall from trigonometry that $\arcsin (\frac{1}{2})=\frac{\pi }{6}$.

$$m=\frac{\pi }{6}+\frac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}=\frac{\pi }{6}+\frac{\frac{1}{2}}{\sqrt{\frac{3}{4}}}=\frac{\pi }{6}+\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$

$$m=\frac{\pi }{6}+\frac{1}{2}\cdot \frac{2}{\sqrt{3}}=\frac{\pi }{6}+\frac{1}{\sqrt{3}}$$

Step 3: Find the y-coordinate of the point of tangency by evaluating $g(\frac{1}{2})$.

$$y_1=g\left(\frac{1}{2}\right)=\frac{1}{2}\cdot \arcsin \left(\frac{1}{2}\right)=\frac{1}{2}\cdot \frac{\pi }{6}=\frac{\pi }{12}$$

The point of tangency is $(\frac{1}{2},\frac{\pi }{12})$.

Step 4: Write the equation of the tangent line using point-slope form, $y-y_1=m(x-x_1)$.

$$y-\frac{\pi }{12}=\left(\frac{\pi }{6}+\frac{1}{\sqrt{3}}\right)\left(x-\frac{1}{2}\right)$$

This is a complete and acceptable final answer on the AP Exam.

Using Your Calculator

This topic is primarily analytical, meaning you are expected to find the derivative formula by hand. A calculator cannot produce the symbolic derivative (e.g., it cannot turn $\arctan (4x^2)$ into \frac{8x}{1+16x^4}`). However, a calculator is an excellent tool for **checking your work** at a specific point. **To verify the slope in Example 2:** 1. On a TI-84 style calculator, press `MATH` and select `nDeriv(` (or press `ALPHA` `WINDOW` and select $d/dx( on newer models).

2. Enter the expression, the variable, and the point at which you want to evaluate the derivative. The syntax is: nDeriv(function, X, value).

3. For Example 2, you would enter: nDeriv(X*sin⁻¹(X), X, 1/2).

   • Note: $si{n}^{-1}($ is typically found by pressing 2nd$SIN$.

4. The calculator will return a decimal approximation, $\approx 1.1027$.

5. Now, calculate the decimal approximation of your analytical answer: $m=\frac{\pi }{6}+\frac{1}{\sqrt{3}}\approx 0.5236+0.5774\approx 1.101$.

6. The values are extremely close (differing only due to calculator rounding), confirming that your derivative calculation is likely correct.

AP Exam Quick Hit

Common Question Types

• Direct Chain Rule Application: A multiple-choice question asking for the derivative of a function like $f(x)=\arcsin (e^{3x})$. This tests your ability to apply the chain rule with an inverse trig function.

• Finding the Equation of a Tangent Line: A free-response or multiple-choice question asking for the tangent line to a function involving an inverse trig function at a given point, as shown in Example 2. This combines differentiation with geometric understanding.

• Combined with Product/Quotient Rule: A multiple-choice question where the inverse trig function is one part of a larger product or quotient, such as finding the derivative of $y=\frac{\arctan (x)}{x^2+1}$.

Common Mistakes

• Forgetting the Chain Rule: This is the most frequent error. Students will differentiate $\arctan (u)$ as $\frac{1}{1+u^2}$ but completely forget to multiply by the derivative of $u$.

• Incorrectly Squaring the Inner Function: When finding the derivative of $\arctan (u)$, a common algebraic mistake is to improperly calculate $u^2$. For example, if $u=5x$, students might write $5x^2$ instead of the correct $(5x{)}^2=25x^2$.

• Mixing Up Formulas: Students often confuse the formulas for $\arcsin (x)$ and $\arctan (x)$, for instance, by putting a square root in the denominator of the $\arctan$ derivative or forgetting it for the $\arcsin$ derivative.

• Sign Errors: Forgetting the negative sign in the derivative of $\arccos (x)$.

• Notation Misinterpretation: Treating ${\sin }^{-1}(x)$ as $(\sin (x){)}^{-1}=\csc (x)$. Remember, the $-1$ superscript denotes an inverse function, not a reciprocal.