Differentiating Inverse Trigonometric Functions - AP Calculus BC Study Guide

The Core Idea: Differentiating Inverse Trigonometric Functions

This topic introduces the rules for finding the derivatives of the six inverse trigonometric functions: $arcsin(x)$, $arccos(x)$, $arctan(x)$, $arccot(x)$, $arcsec(x)$, and $arccsc(x)$. Just as we have specific rules for differentiating standard trigonometric, polynomial, and exponential functions, we must also have a set of rules for their inverses. The core task is to calculate the instantaneous rate of change of these functions, which often model angles as a function of a ratio.

A critical component of this topic is its connection to the chain rule. Rarely on the AP Exam will you be asked to differentiate just $arcsin(x)$. Instead, you will typically need to differentiate a composite function, such as $arcsin(u(x))$, where $u(x)$ is another function. Therefore, mastering the application of the chain rule in conjunction with the six new derivative formulas is the fundamental skill required by this topic.

Key Formulas

The following are the required derivative formulas for the six inverse trigonometric functions. It is essential to commit these to memory. Note that the notation $arcsin(x)$ is equivalent to $si{n}^{-1}(x)$. The "arc" notation is often preferred to avoid confusion with the reciprocal, $(sin(x){)}^{-1}$.

• Derivative of Arcsine

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

• Derivative of Arccosine

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

• Derivative of Arctangent

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

• Derivative of Arccotangent

  $$\frac{d}{dx}(\text{arccot}(x))=-\frac{1}{1+x^2}$$

• Derivative of Arcsecant

  $$\frac{d}{dx}(\text{arcsec}(x))=\frac{1}{|x|\sqrt{x^2-1}}$$

• Derivative of Arccosecant

  $$\frac{d}{dx}(\text{arccsc}(x))=-\frac{1}{|x|\sqrt{x^2-1}}$$

Understanding the Chain Rule with Inverse Trig Functions

The Essential Knowledge for this topic explicitly states that the chain rule must be applied to find the derivatives of compositions involving inverse trigonometric functions. This means that the argument of the inverse trigonometric function will often be a function $u(x)$ rather than simply $x$.

The key is to generalize the basic formulas using the chain rule. If $u$ is a differentiable function of $x$, the rules become:

• $$\frac{d}{dx}(\arcsin (u))=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$$

• $$\frac{d}{dx}(\arccos (u))=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$$

• $$\frac{d}{dx}(\arctan (u))=\frac{1}{1+u^2}\cdot \frac{du}{dx}$$

• $$\frac{d}{dx}(\text{arccot}(u))=-\frac{1}{1+u^2}\cdot \frac{du}{dx}$$

• $$\frac{d}{dx}(\text{arcsec}(u))=\frac{1}{|u|\sqrt{u^2-1}}\cdot \frac{du}{dx}$$

• $$\frac{d}{dx}(\text{arccsc}(u))=-\frac{1}{|u|\sqrt{u^2-1}}\cdot \frac{du}{dx}$$

In every problem, you must first identify the "inner function" $u$, substitute it into the appropriate derivative formula, and then multiply the entire result by the derivative of $u$, which is $du/dx$. This final multiplication step is the application of the chain rule and is the most frequently forgotten step by students.

Core Concepts & Rules

• Six Fundamental Formulas: There are six distinct derivative rules, one for each inverse trigonometric function. These must be memorized.

• Co-function Relationships: The derivatives of co-function pairs are negatives of each other.

  • The derivative of $arccos(x)$ is the negative of the derivative of $arcsin(x)$.

  • The derivative of $arccot(x)$ is the negative of the derivative of $arctan(x)$.

  • The derivative of $arccsc(x)$ is the negative of the derivative of $arcsec(x)$.

• Chain Rule is Paramount: When differentiating an inverse trigonometric function where the input is any expression other than $x$, the chain rule must be used. The derivative of the "inside" function must be multiplied by the result.

• Absolute Value is Required: The formulas for the derivatives of $arcsec(x)$ and $arccsc(x)$ include an absolute value, $|x|$, in the denominator. This is a formal part of the definition and cannot be omitted.

Step-by-Step Example 1: Basic Application

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

Solution:

This problem requires the derivative formula for $arctan(u)$ combined with the chain rule.

Step 1: Identify the outer and inner functions.

• The outer function is $\arctan (u)$.

• The inner function is $u=e^{2x}$.

Step 2: State the derivative rule for the outer function.

The general formula is:

$$\frac{d}{dx}(\arctan (u))=\frac{1}{1+u^2}\cdot \frac{du}{dx}$$

Step 3: Find the derivative of the inner function, $du/dx$.

The inner function is $u=e^{2x}$. Its derivative requires another application of the chain rule.

$$\frac{du}{dx}=\frac{d}{dx}(e^{2x})=e^{2x}\cdot 2=2e^{2x}$$

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

Substitute $u=e^{2x}$ and $du/dx=2e^{2x}$ into the formula from Step 2.

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

Step 5: Simplify the final expression.

Using exponent rules, $(e^{2x}{)}^2=e^{4x}$.

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

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 $x=1/2$.

Solution:

This problem requires finding a point and a slope. The slope is the value of the derivative at the given point. Finding the derivative requires the product rule in combination with the inverse trigonometric derivative rule.

Step 1: Find the point of tangency $(x_1,y_1)$.

We are given $x_1=1/2$. We find $y_1$ by evaluating $g(1/2)$.

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

Recall that $arcsin(1/2)$ is the angle $\theta$ in $[-\pi /2,\pi /2]$ such that $sin(\theta )=1/2$. This angle is $\pi /6$.

$$y_1=\frac{1}{2}\cdot \frac{\pi }{6}=\frac{\pi }{12}$$

The point of tangency is $(1/2,\pi /12)$.

Step 2: Find the derivative of $g(x)$ using the Product Rule.

The function $g(x)$ is a product of two functions: $x$ 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 product 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 3: Evaluate the derivative at $x=1/2$ to find the slope $m$.

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

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

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

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

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

Step 4: Write the equation of the tangent line using point-slope form.

Using the point $(1/2,\pi /12)$ and the slope $m=\pi /6+1/\sqrt{3}$:

$$y-y_1=m(x-x_1)$$

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

Using Your Calculator

The derivatives of inverse trigonometric functions are found analytically using the formulas. A calculator is not used to find the symbolic derivative. However, it is an excellent tool for verifying your answer at a specific point.

To check the slope calculated in Example 2 at $x=1/2$, you can use the numerical differentiation feature of your calculator (e.g., nDeriv on a TI-84 or $d/dx|$ on a TI-Nspire).

Verification Steps:

1. Calculate the exact value of your analytical derivative.

   From Example 2, our slope was $m=\pi /6+1/\sqrt{3}$.

   $\pi /6+1/\sqrt{3}\approx 0.52359+0.57735\approx 1.10094$.

2. Use the calculator's numerical derivative function on the original function.

   On the home screen, input the command to find the derivative of $g(x)=x\cdot \arcsin (x)$ at $x=1/2$.

   • TI-84 Syntax: nDeriv(X*sin⁻¹(X), X, 1/2)

   • TI-Nspire Syntax: $d/dx(x\ast si{n}^{-1}(x))|x=1/2$

3. Compare the results.

   The calculator will return a value approximately equal to $1.10094$. Since this matches the decimal approximation of our exact analytical answer, we can be confident our derivative calculation is correct.

AP Exam Quick Hit

Common Question Types

• Direct Differentiation with Chain Rule: This is the most common format. You will be given a function that is a composition and asked to find its derivative.

  • Example: Find $dy/dx$ if $y=\text{arcsec}(4x^3)$.

• Combination with Other Rules: Questions often require you to use the product rule, quotient rule, or implicit differentiation in conjunction with an inverse trigonometric derivative.

  • Example: Find the derivative of $f(x)=\frac{\arctan (x)}{x^2+1}$.

• Finding Slopes and Tangent Lines: You will be asked to find the slope of a function at a point, or the full equation of a tangent line, which requires evaluating the derivative.

  • Example: Find the slope of the tangent line to $f(x)=\arccos (x)$ at $x=-\sqrt{2}/2$.

Common Mistakes

• Forgetting the Chain Rule: The most frequent error is differentiating the outer inverse trigonometric function but forgetting to multiply by the derivative of the inner function $u(x)$. For $arctan(e^{2x})$, a common error is to write $1/(1+(e^{2x}{)}^2)$ and forget to multiply by $2e^{2x}$.

• Incorrect Formula Memorization: Students often mix up the formulas, particularly the signs. A common mistake is forgetting the negative sign on the derivatives of $arccos(x)$, $arccot(x)$, and $arccsc(x)$.

• Omitting the Absolute Value: Forgetting the absolute value in the denominator for the derivatives of $arcsec(u)$ and $arccsc(u)$ is a critical error. The formula is $1/(|u|\sqrt{u^2-1})$, not $1/(u\sqrt{u^2-1})$.

• Algebraic Simplification Errors: After correctly applying the chain rule, students often make mistakes simplifying the resulting expression. For example, when finding the derivative of $arcsin(3x)$, a student might correctly write $\frac{1}{\sqrt{1-(3x{)}^2}}\cdot 3$ but incorrectly simplify the denominator to $\sqrt{1-3x^2}$ instead of the correct $\sqrt{1-9x^2}$.