#AP Calculus AB/BC: Derivatives of Inverse Trigonometric Functions

Hey there, future AP Calculus master! 👋 Let's dive into the world of inverse trig derivatives. This guide is your fast track to acing those tricky problems. We'll break it down, make it stick, and get you feeling confident for test day!

#Understanding Inverse Trig Derivatives

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Remember, inverse trig functions 'undo' what regular trig functions do. So, if $\sin(y) = x$, then $y = \sin^{-1}(x)$. We're going to find out how to differentiate these inverse functions.

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The golden rule for finding the derivative of an inverse function is:

$$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$$

This might look scary, but it's just saying: "The derivative of the inverse is one over the derivative of the original function, evaluated at the inverse function." Let's see how it works with $\sin^{-1}(x)$.

#Derivative of $\sin^{-1}(x)$

Let's find the derivative of $y = \sin^{-1}(x)$.

1. Start with the Inverse: If $y = \sin^{-1}(x)$, then $x = \sin(y)$.
2. Implicit Differentiation: Differentiate both sides with respect to $x$.
   $$1 = \cos(y) \frac{dy}{dx}$$
3. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{\cos(y)}$$
4. Rewrite in terms of x: We know $\sin(y) = x$. Use the identity $\sin^2(y) + \cos^2(y) = 1$ to find $\cos(y)$. $$\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2}$$
5. Final Answer: $$\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}$$

#The Full List of Inverse Trig Derivatives

Here's the cheat sheet you'll need for the exam. Memorize these!

#Practice Makes Perfect

Let's put these derivatives to work! Remember to use the chain rule when the inside function is not just a simple 'x'.

#Example 1: $\sin^{-1}(3x)$

If $y = \sin^{-1}(3x)$, find $\frac{dy}{dx}$.

1. Apply the Chain Rule: $$\frac{dy}{dx} = \frac{1}{\sqrt{1-(3x)^2}} \cdot \frac{d}{dx}(3x)$$
2. Simplify: $$\frac{dy}{dx} = \frac{3}{\sqrt{1-9x^2}}$$

#Example 2: $\tan^{-1}(x+6)$

If $y = \tan^{-1}(x+6)$, find $\frac{dy}{dx}$.

1. Apply the Chain Rule: $$\frac{dy}{dx} = \frac{1}{1+(x+6)^2} \cdot \frac{d}{dx}(x+6)$$
2. Simplify: $$\frac{dy}{dx} = \frac{1}{1+(x^2+12x+36)} = \frac{1}{x^2+12x+37}$$

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• Forgetting the Chain Rule: Always remember to multiply by the derivative of the inside function.
• Incorrect Simplification: Be careful when simplifying expressions under square roots or in denominators.
• Mixing up Derivatives: Double-check you're using the correct derivative for each inverse trig function.

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Practice Question

Practice Questions

#Multiple Choice Questions

1. What is the derivative of $f(x) = \cos^{-1}(2x)$?

   (A) $\frac{1}{\sqrt{1-4x^2}}$ (B) $\frac{-1}{\sqrt{1-4x^2}}$ (C) $\frac{2}{\sqrt{1-4x^2}}$ (D) $\frac{-2}{\sqrt{1-4x^2}}$

2. If $g(x) = \tan^{-1}(x^2)$, what is $g'(x)$?

   (A) $\frac{1}{1+x^4}$ (B) $\frac{2x}{1+x^4}$ (C) $\frac{1}{1+x^2}$ (D) $\frac{2x}{1+x^2}$

#Free Response Question

Let $h(x) = x \cdot \sin^{-1}(x)$.

(a) Find $h'(x)$.

(b) Find the equation of the tangent line to $h(x)$ at $x = \frac{1}{2}$.

(c) Evaluate $\lim_{x \to 0} \frac{h(x)}{x}$.

#Solutions

Multiple Choice:

1. Correct Answer: (D) $\frac{d}{dx} [\cos^{-1}(2x)] = -\frac{1}{\sqrt{1-(2x)^2}} \cdot 2 = \frac{-2}{\sqrt{1-4x^2}}$

2. Correct Answer: (B) $g'(x) = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4}$

Free Response:

(a) [3 points]

• Use the product rule: $h'(x) = (1) \cdot \sin^{-1}(x) + x \cdot \frac{1}{\sqrt{1-x^2}}$
• $h'(x) = \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$

(b) [3 points]

• Find $h(\frac{1}{2}) = \frac{1}{2} \sin^{-1}(\frac{1}{2}) = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12}$
• Find $h'(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) + \frac{\frac{1}{2}}{\sqrt{1-(\frac{1}{2})^2}} = \frac{\pi}{6} + \frac{\frac{1}{2}}{\sqrt{\frac{3}{4}}} = \frac{\pi}{6} + \frac{1}{\sqrt{3}} = \frac{\pi}{6} + \frac{\sqrt{3}}{3}$
• Tangent line: $y - \frac{\pi}{12} = (\frac{\pi}{6} + \frac{\sqrt{3}}{3})(x - \frac{1}{2})$

(c) [3 points]

• Use L'Hôpital's Rule: $\lim_{x \to 0} \frac{h(x)}{x} = \lim_{x \to 0} \frac{h'(x)}{1}$
• $\lim_{x \to 0} h'(x) = \lim_{x \to 0} \left( \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}} \right) = \sin^{-1}(0) + \frac{0}{\sqrt{1-0}} = 0$
• $\lim_{x \to 0} \frac{h(x)}{x} = 0$

#Final Exam Focus

• Memorize the Derivatives: Know them cold! This is a must for both MCQs and FRQs.
• Chain Rule is Key: Inverse trig functions are often part of composite functions, so practice using the chain rule.
• Implicit Differentiation: Be ready to use implicit differentiation to derive these formulas yourself if needed.
• Applications: Expect to see these derivatives in related rates, optimization, and curve sketching problems.

#Last-Minute Tips

• Time Management: Don't spend too long on any one question. If you're stuck, move on and come back later.
• Show Your Work: Even if you make a mistake, you can get partial credit if you show your steps.
• Stay Calm: You've got this! Take deep breaths and focus on what you know.

You're now equipped with the knowledge and strategies to tackle inverse trig derivatives. Go get 'em! 💪