#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}
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