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Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Abigail Young

Abigail Young

6 min read

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Study Guide Overview

This study guide covers derivatives of advanced trigonometric functions, specifically tan(x), cot(x), sec(x), and csc(x). It provides a quick reference table of the derivatives, detailed examples, mnemonics, common mistakes, practice problems, and final exam tips. The guide emphasizes the importance of radians, the chain rule, and simplifying trigonometric expressions before differentiating.

AP Calculus AB/BC: Derivatives of Advanced Trigonometric Functions 🚀

Hey there, future calculus champ! Let's get these advanced trig derivatives locked down. This guide is designed to be your go-to resource for a quick, effective review. Let's make sure you're feeling confident and ready to ace the exam!

Derivatives of tan(x), cot(x), sec(x), and csc(x)

Quick Reference Table

FunctionDerivative
f(x)=tanxf(x) = \tan xf(x)=sec2xf'(x) = \sec^2 x
g(x)=cotxg(x) = \cot xg(x)=csc2xg'(x) = -\csc^2 x
h(x)=secxh(x) = \sec xh(x)=secxtanxh'(x) = \sec x \tan x
k(x)=cscxk(x) = \csc xk(x)=cscxcotxk'(x) = -\csc x \cot x
Quick Fact

Remember: These rules apply only when angles are in radians, not degrees!

Memory Aid

Mnemonic Alert!

  • Tan becomes secant squared (tanxsec2x\tan x \rightarrow \sec^2 x)
  • Co-functions (cot, csc) always have a negative derivative.
  • Secant and cosecant derivatives involve both secant/cosecant and tangent/cotangent.

Key Concept

Individual Derivatives: A Closer Look

Derivative of tanx\tan x

The derivative of tanx\tan x is sec2x\sec^2 x. Let's see it in action:

Example: f(x)=3tanx+2x2f(x) = 3\tan x + 2x^2

To find f(x)f'(x), differentiate each term separately:

  • Derivative of 3tanx3\tan x is 3sec2x3\sec^2 x.
  • Derivative of 2x22x^2 is 4x4x.

Thus, f(x)=3sec2x+4xf'(x) = 3\sec^2 x + 4x

Derivative of cotx\cot x

The derivative of cotx\cot x is csc2x-\csc^2 x. Check out this example:

Example: f(x)=5cotx+xf(x) = 5\cot x + x

  • Derivative of 5cotx5\cot x is 5csc2x-5\csc^2 x.
  • Derivative of xx is 11.

Therefore, f(x)=5csc2x+1f'(x) = -5\csc^2 x + 1

Derivative of secx\sec x

The derivative of secx\sec x is secxtanx\sec x \tan x. Let's see an example:

Example: f(x)=2secx+3x3f(x) = 2\sec x + 3x^3

  • Derivative of 2secx2\sec x is 2secxtanx2\sec x \tan x.
  • Derivative of 3x33x^3 is 9x29x^2.

So, f(x)=2secxtanx+9x2f'(x) = 2\sec x \tan x + 9x^2

Derivative of cscx\csc x

The derivative of cscx\csc x is cscxcotx-\csc x \cot x. Here's an example:

Example: f(x)=4cscx+7x2f(x) = 4\csc x + 7x^2

  • Derivative of 4cscx4\csc x is 4cscxcotx-4\csc x \cot x.
  • Derivative of 7x27x^2 is 14x14x.

Therefore, f(x)=4cscxcotx+14xf'(x) = -4\csc x \cot x + 14x

Exam Tip

Simplify First! Before taking derivatives, use trig identities to simplify complex expressions. For example, remember that tan(x)=sin(x)cos(x)\tan(x) = \frac{\sin(x)}{\cos(x)} and cot(x)=1tan(x)\cot(x) = \frac{1}{\tan(x)}.

🏋️‍♂️ Practice Problems

Let's solidify your understanding with some practice! Remember to apply the chain rule, sum rule, and quotient rule as needed.

❓ Advanced Trig Derivative Practice Questions

Find the derivatives for the following:

  1. f(x)=2tan(x)+sec(x)f(x) = 2 \tan(x) + \sec(x)
  2. f(x)=cot(x)csc(x)f(x) = \frac{\cot(x)}{\csc(x)}
  3. g(x)=tan2(6x)g(x) = \tan^2(6x)
  4. h(x)=5cot(x)h(x) = 5\cot(x)

🤔 Advanced Trig Derivative Practice Solutions

  1. f(x)=2sec2(x)+sec(x)tan(x)f'(x) = 2 \sec^2(x) + \sec(x) \tan(x)
  2. f(x)=csc2(x)f'(x) = -\csc^2(x)
  3. g(x)=12tan(6x)sec2(6x)g'(x) = 12\tan(6x)\sec^2(6x) (Chain rule!)
  4. h(x)=5csc2(x)h'(x) = -5\csc^2(x)
Common Mistake

Watch Out! Don't forget the chain rule when differentiating composite functions (like tan2(6x)\tan^2(6x)). Also, double-check your signs, especially with co-functions.

🎯 Final Exam Focus

  • High-Priority Topics: Derivatives of tanx\tan x, cotx\cot x, secx\sec x, and cscx\csc x are frequently tested, often combined with other derivative rules.
  • Common Question Types: Expect to see these derivatives in both multiple-choice and free-response questions. Be prepared to apply them within the context of related rates, optimization, and curve sketching problems.
  • Time Management: Practice makes perfect! The more you practice, the faster you'll become at recognizing and applying these rules.
  • Pitfalls: Be careful with signs (especially negative signs with co-functions) and remember to apply the chain rule when needed.

Practice Question

Practice Questions

Multiple Choice Questions

  1. What is the derivative of f(x)=3sec(x)2tan(x)f(x) = 3\sec(x) - 2\tan(x)? (A) 3sec(x)tan(x)2sec2(x)3\sec(x)\tan(x) - 2\sec^2(x) (B) 3csc(x)cot(x)2cot2(x)-3\csc(x)\cot(x) - 2\cot^2(x) (C) 3sec(x)tan(x)+2sec2(x)3\sec(x)\tan(x) + 2\sec^2(x) (D) 3sec2(x)2sec(x)tan(x)3\sec^2(x) - 2\sec(x)\tan(x)

  2. If g(x)=cot(x)xg(x) = \frac{\cot(x)}{x}, what is g(x)g'(x)? (A) xcsc2(x)cot(x)x2\frac{-x\csc^2(x) - \cot(x)}{x^2} (B) xcsc2(x)cot(x)x2\frac{x\csc^2(x) - \cot(x)}{x^2} (C) csc2(x)1\frac{-\csc^2(x)}{1} (D) xcsc2(x)+cot(x)x2\frac{-x\csc^2(x) + \cot(x)}{x^2}

Free Response Question

Consider the function h(x)=2csc(x)+tan(x)h(x) = 2\csc(x) + \tan(x).

(a) Find h(x)h'(x). (b) Find the equation of the tangent line to h(x)h(x) at x=π4x = \frac{\pi}{4}. (c) Determine all values of xx in the interval [0,2π][0, 2\pi] where h(x)=0h'(x) = 0.

Scoring Rubric

(a) 2 points

  • 1 point for correctly differentiating 2csc(x)2\csc(x) as 2csc(x)cot(x)-2\csc(x)\cot(x).
  • 1 point for correctly differentiating tan(x)\tan(x) as sec2(x)\sec^2(x).

(b) 3 points

  • 1 point for finding the correct slope h(π4)h'(\frac{\pi}{4}).
  • 1 point for finding the correct y-value h(π4)h(\frac{\pi}{4}).
  • 1 point for writing the tangent line equation in point-slope form.

(c) 4 points

  • 1 point for setting h(x)=0h'(x) = 0.
  • 2 points for correctly solving the trigonometric equation.
  • 1 point for identifying all correct solutions in the interval [0,2π][0, 2\pi].

Answers:

Multiple Choice:

  1. (A)
  2. (A)

Free Response:

(a) h(x)=2csc(x)cot(x)+sec2(x)h'(x) = -2\csc(x)\cot(x) + \sec^2(x) (b) h(π4)=22+2h'(\frac{\pi}{4}) = -2\sqrt{2} + 2, h(π4)=22+1h(\frac{\pi}{4}) = 2\sqrt{2} + 1. Tangent line: y(22+1)=(22+2)(xπ4)y - (2\sqrt{2} + 1) = (-2\sqrt{2} + 2)(x - \frac{\pi}{4}) (c) x=π2,3π2x = \frac{\pi}{2}, \frac{3\pi}{2}

🌟 Closing

You've got this! Keep practicing, and you'll be acing those derivative problems in no time. Remember, every step you take is a step closer to success. Good luck, and happy calculating! 🌈

Question 1 of 11

What is the derivative of f(x)=2sec(x)f(x) = 2\sec(x)? 🚀

2sec(x)tan(x)2\sec(x)\tan(x)

2csc(x)cot(x)-2\csc(x)\cot(x)

2sec2(x)2\sec^2(x)

2sec(x)tan(x)-2\sec(x)\tan(x)