The derivative of the inverse sine function, \( \sin^{-1}(x) \), with respect to \(x\) is:

\[
\frac{d}{dx} \left( \sin^{-1}(x) \right) = \frac{1}{\sqrt{1 - x^2}}
\]

This derivative is valid for \( x \) in the interval \(-1 \leq x \leq 1\).

Would you like more details on how this derivative is derived, or do you have any questions?

Here are 5 related questions you might find interesting:
1. What is the derivative of \( \cos^{-1}(x) \)?
2. How do you find the derivative of the inverse tangent function \( \tan^{-1}(x) \)?
3. Can you derive the derivative of \( \sin^{-1}(x) \) using implicit differentiation?
4. What is the integral of \( \frac{1}{\sqrt{1 - x^2}} \)?
5. How does the domain of \( \sin^{-1}(x) \) affect its derivative?

**Tip:** Remember that the derivatives of inverse trigonometric functions often involve square roots and are derived by considering the right triangles and their corresponding trigonometric identities.

Learn how to find the derivative of \( \sin^{-1}(x) \) step-by-step with detailed explanations and examples. Understand the formula \( \frac{d}{dx} \left( \sin^{-1}(x) \right) = \frac{1}{\sqrt{1 - x^2}} \) and its application. Suitable for advanced high school students and anyone interested in calculus and trigonometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation