#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

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#Derivative of $\tan x$

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

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

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

• Derivative of $3\tan x$ is $3\sec^2 x$.
• Derivative of $2x^2$ is $4x$.

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

#Derivative of $\cot x$

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

Example: $$f(x) = 5\cot x + x$$

• Derivative of $5\cot x$ is $-5\csc^2 x$.
• Derivative of $x$ is $1$.

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

#Derivative of $\sec x$

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

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

• Derivative of $2\sec x$ is $2\sec x \tan x$.
• Derivative of $3x^3$ is $9x^2$.

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

#Derivative of $\csc x$

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

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

• Derivative of $4\csc x$ is $-4\csc x \cot x$.
• Derivative of $7x^2$ is $14x$.

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

#🏋️‍♂️ 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. $f(x) = 2 \tan(x) + \sec(x)$
2. $f(x) = \frac{\cot(x)}{\csc(x)}$
3. $g(x) = \tan^2(6x)$
4. $h(x) = 5\cot(x)$

#🤔 Advanced Trig Derivative Practice Solutions

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

#🎯 Final Exam Focus

• High-Priority Topics: Derivatives of $\tan x$, $\cot x$, $\sec x$, and $\csc x$ 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.

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Practice Question

Practice Questions

#Multiple Choice Questions

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

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

#Free Response Question

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

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

Scoring Rubric

(a) 2 points

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

(b) 3 points

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

(c) 4 points

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

Answers:

Multiple Choice:

1. (A)
2. (A)

Free Response:

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

#🌟 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! 🌈