Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions - AP Calculus AB Study Guide

The Core Idea: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

This topic expands our differentiation toolkit to include all six standard trigonometric functions. While the derivatives of sine and cosine form the foundation of trigonometric calculus, many real-world models and mathematical problems involve tangent, cotangent, secant, and cosecant. The central concept is that we do not need to discover these new derivative rules from scratch using the limit definition.

Instead, the derivatives of these four functions can be logically derived using rules we already know. By expressing tangent, cotangent, secant, and cosecant in terms of sine and cosine, we can apply the quotient rule (and sometimes the product rule) to find their derivatives. This process reinforces the interconnectedness of calculus rules and demonstrates how complex derivatives can be built from simpler, foundational ones. The ultimate goal is to both understand these derivations and commit the final derivative formulas to memory for efficient problem-solving.

Key Formulas

The derivatives of the six trigonometric functions are essential for calculus. The first two are foundational, and the remaining four are derived from them.

• Derivative of Sine:

  $$\frac{d}{dx}(\sin (x))=\cos (x)$$

• Derivative of Cosine:

  $$\frac{d}{dx}(\cos (x))=-\sin (x)$$

• Derivative of Tangent:

  $$\frac{d}{dx}(\tan (x))={\sec }^2(x)$$

• Derivative of Cotangent:

  $$\frac{d}{dx}(\cot (x))=-{\csc }^2(x)$$

• Derivative of Secant:

  $$\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$$

• Derivative of Cosecant:

  $$\frac{d}{dx}(\csc (x))=-\csc (x)\cot (x)$$

Understanding the Derivations

A critical piece of essential knowledge is that the derivatives for tangent, cotangent, secant, and cosecant are not arbitrary; they are the direct result of applying the quotient rule to the definitions of these functions in terms of sine and cosine. Understanding these derivations is key to deeper comprehension and can help you re-create a formula if you forget it.

Deriving the Derivative of Tangent

The tangent function is defined as $\tan (x)=\frac{\sin (x)}{\cos (x)}$. To find its derivative, we apply the quotient rule, $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g{f}^{\prime }-f{g}^{\prime }}{g^2}$, where $f(x)=\sin (x)$ and $g(x)=\cos (x)$.

1. Identify $f(x)$ and $g(x)$:

   • $f(x)=\sin (x)$

   • $g(x)=\cos (x)$

2. Find their derivatives:

   • $f^{\prime }(x)=\cos (x)$

   • $g^{\prime }(x)=-\sin (x)$

3. Apply the quotient rule formula:

   $$\frac{d}{dx}(\tan (x))=\frac{(\cos (x))(\cos (x))-(\sin (x))(-\sin (x))}{(\cos (x){)}^2}$$

4. Simplify the numerator:

   $$\frac{d}{dx}(\tan (x))=\frac{{\cos }^2(x)+{\sin }^2(x)}{{\cos }^2(x)}$$

5. Apply the Pythagorean Identity:

   Recall the fundamental identity ${\sin }^2(x)+{\cos }^2(x)=1$.

   $$\frac{d}{dx}(\tan (x))=\frac{1}{{\cos }^2(x)}$$

6. Use the reciprocal identity:

   Since $\sec (x)=\frac{1}{\cos (x)}$, it follows that ${\sec }^2(x)=\frac{1}{{\cos }^2(x)}$.

   $$\frac{d}{dx}(\tan (x))={\sec }^2(x)$$

Deriving the Derivative of Secant

The secant function is defined as $\sec (x)=\frac{1}{\cos (x)}$. We can again use the quotient rule with $f(x)=1$ and $g(x)=\cos (x)$.

1. Identify $f(x)$ and $g(x)$:

   • $f(x)=1$

   • $g(x)=\cos (x)$

2. Find their derivatives:

   • $f^{\prime }(x)=0$ (The derivative of a constant is zero)

   • $g^{\prime }(x)=-\sin (x)$

3. Apply the quotient rule formula:

   $$\frac{d}{dx}(\sec (x))=\frac{(\cos (x))(0)-(1)(-\sin (x))}{(\cos (x){)}^2}$$

4. Simplify the expression:

   $$\frac{d}{dx}(\sec (x))=\frac{0+\sin (x)}{{\cos }^2(x)}=\frac{\sin (x)}{{\cos }^2(x)}$$

5. Rewrite and separate the fraction:

   To get the final form, we can rewrite the fraction as a product.

   $$\frac{d}{dx}(\sec (x))=\frac{1}{\cos (x)}\cdot \frac{\sin (x)}{\cos (x)}$$

6. Use the reciprocal and quotient identities:

   Recognize that $\frac{1}{\cos (x)}=\sec (x)$ and $\frac{\sin (x)}{\cos (x)}=\tan (x)$.

   $$\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$$

The derivations for $\cot (x)$ and $\csc (x)$ follow a very similar process.

Core Concepts & Rules

• Memorization is Key: The six trigonometric derivative formulas must be memorized for the AP Exam. Fluency with these rules is expected.

• Foundation in Sine and Cosine: All four "other" trigonometric derivatives ($\tan (x)$, $\cot (x)$, $\sec (x)$, $\csc (x)$) are derived from the derivatives of $\sin (x)$ and $\cos (x)$.

• Quotient Rule is the Tool: The primary method for deriving these formulas is the quotient rule, applied to the definitions of the functions (e.g., $\tan (x)=\sin (x)/\cos (x)$).

• Pattern Recognition: Notice the patterns in the derivatives. The derivatives of the "co-" functions ($\cos (x)$, $\cot (x)$, $\csc (x)$) are all negative. This can serve as a quick mental check.

• Pythagorean Identities are Crucial: Simplifying the results of the quotient rule often requires using the Pythagorean identity ${\sin }^2(x)+{\cos }^2(x)=1$.

Step-by-Step Example 1: Basic Application

Problem: Find the derivative of the function $f(x)=5\tan (x)-\csc (x)$.

Solution:

This problem requires applying the constant multiple rule, the difference rule, and the new derivative rules for tangent and cosecant.

Step 1: Differentiate term by term.

The derivative of a difference is the difference of the derivatives.

$$f^{\prime }(x)=\frac{d}{dx}(5\tan (x))-\frac{d}{dx}(\csc (x))$$

Step 2: Apply the constant multiple rule to the first term.

The constant $5$ can be factored out of the derivative.

$$f^{\prime }(x)=5\cdot \frac{d}{dx}(\tan (x))-\frac{d}{dx}(\csc (x))$$

Step 3: Apply the specific trigonometric derivative rules.

We use the memorized rules: $\frac{d}{dx}(\tan (x))={\sec }^2(x)$ and $\frac{d}{dx}(\csc (x))=-\csc (x)\cot (x)$.

$$f^{\prime }(x)=5({\sec }^2(x))-(-\csc (x)\cot (x))$$

Step 4: Simplify the expression.

The double negative becomes a positive.

$$f^{\prime }(x)=5{\sec }^2(x)+\csc (x)\cot (x)$$

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 $y=\sec (x)$ at the point where $x=\frac{\pi }{3}$.

Solution:

To find the equation of a tangent line, we need two things: a point on the line and the slope of the line at that point. The slope is found by evaluating the derivative at the given x-value.

Step 1: Find the point of tangency.

We are given the x-coordinate, $x=\frac{\pi }{3}$. We need to find the corresponding y-coordinate by plugging this value into the original function.

$$y=\sec \left(\frac{\pi }{3}\right)$$

Recall that $\sec (x)=\frac{1}{\cos (x)}$.

$$y=\frac{1}{\cos \left(\frac{\pi }{3}\right)}=\frac{1}{1/2}=2$$

So, the point of tangency is $(\frac{\pi }{3},2)$.

Step 2: Find the derivative of the function.

The derivative of $y=\sec (x)$ is the slope formula for the tangent line at any point $x$.

$$\frac{dy}{dx}=\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$$

Step 3: Evaluate the derivative at the given x-value to find the slope.

We plug $x=\frac{\pi }{3}$ into the derivative.

$$m=\frac{dy}{dx}{|}_{x=\pi /3}=\sec \left(\frac{\pi }{3}\right)\tan \left(\frac{\pi }{3}\right)$$

We already found $\sec (\frac{\pi }{3})=2$. We also know $\tan (\frac{\pi }{3})=\sqrt{3}$.

$$m=(2)(\sqrt{3})=2\sqrt{3}$$

The slope of the tangent line at $x=\frac{\pi }{3}$ is $2\sqrt{3}$.

Step 4: Use the point-slope form to write the equation of the line.

The point-slope form is $y-y_1=m(x-x_1)$.

Using our point $(\frac{\pi }{3},2)$ and slope $m=2\sqrt{3}$:

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

This is a perfectly acceptable final answer on the AP Exam.

Using Your Calculator

The derivative rules for tangent, cotangent, secant, and cosecant must be learned and applied analytically (by hand). The calculator is not used to find these general derivative formulas.

However, a graphing calculator can be an excellent tool for checking your work on a problem that asks for the value of a derivative at a specific point.

Example: Check the slope found in Example 2, where we found the derivative of $y=\sec (x)$ at $x=\frac{\pi }{3}$ to be $2\sqrt{3}$.

TI-84 Style Commands:

1. Press the [MATH] button and select [8: nDeriv(] or press [ALPHA][WINDOW] and select [3: nDeriv(].

2. The syntax is $nDeriv(function,variable,value)$.

3. Enter the function. Note that there is no secant button, so you must enter it as $1/cos(X)$.

4. The command will look like this: nDeriv(1/cos(X), X, π/3)

5. Press [ENTER]. The calculator will return a decimal approximation.

   nDeriv(1/cos(X), X, π/3) ≈ 3.464101615

6. Now, calculate the decimal value of your analytical answer, $2\sqrt{3}$.

   $2\ast \sqrt{3}\approx 3.464101615$

7. Since the decimal values match, you can be confident that your analytical derivative calculation is correct.

AP Exam Quick Hit

Common Question Types

• Direct Derivative with Other Rules: You will be asked to find the derivative of a function that combines these new trig functions with the sum, product, quotient, and chain rules.

  • Example: Find $\frac{dy}{dx}$ if $y=e^x\cot (x)$. (This would require the product rule).

• Finding the Slope or Equation of a Tangent Line: A classic application of the derivative. You'll be given a function and a point (or x-value) and asked for the equation of the tangent line.

  • Example: Find the slope of the line normal to the graph of $f(x)=\tan (x)$ at $x=\frac{\pi }{4}$.

• Finding Horizontal Tangents: You may be asked to find the x-values where a function has a horizontal tangent line. This involves setting the derivative equal to zero and solving for x.

  • Example: Find all values of $x$ on the interval $(0,2\pi )$ for which $f(x)=x-\sec (x)$ has a horizontal tangent.

Common Mistakes

• Sign Errors: The most frequent mistake is forgetting the negative sign on the derivatives of the "co-" functions: $\cos (x)$, $\cot (x)$, and $\csc (x)$.

• Formula Mix-ups: Students often confuse the formulas, especially $\frac{d}{dx}(\tan (x))={\sec }^2(x)$ and $\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$. Create flashcards or use mnemonic devices to keep them straight.

• Incorrectly Applying the Chain Rule: When differentiating a composite function like $\tan (4x)$, a common error is to write ${\sec }^2(4x)$ and forget to multiply by the derivative of the inner function, $4$. The correct derivative is $4{\sec }^2(4x)$.

• Algebraic Simplification Errors: When deriving the formulas using the quotient rule, simple algebraic mistakes (like distributing a negative sign incorrectly) can lead to the wrong result.

• Unit Circle Errors: When evaluating a derivative at a specific value, such as $\pi /3$ or $\pi /4$, students may use incorrect values for sine, cosine, or tangent from the unit circle. Reviewing these values is essential.