Prove That The Derivative Of The Function $ Y = \sin^{-1} X $ Is $ Y' = \frac{1}{\sqrt{1-x^2}} $.

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Introduction

The inverse sine function, denoted as $ \sin^{-1} x $, is a fundamental concept in mathematics, particularly in calculus. It is defined as the inverse of the sine function, which means that it returns the angle whose sine is a given value. In this article, we will prove that the derivative of the inverse sine function is $ y' = \frac{1}{\sqrt{1-x^2}} $.

The Inverse Sine Function

The inverse sine function is defined as:

y=sin⁡−1x y = \sin^{-1} x

This function is the inverse of the sine function, which is defined as:

sin⁡y=x \sin y = x

The inverse sine function is a many-to-one function, meaning that there are multiple values of $ y $ that correspond to a single value of $ x $. To resolve this ambiguity, we restrict the domain of the inverse sine function to the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Derivative of the Inverse Sine Function

To find the derivative of the inverse sine function, we can use the chain rule and the fact that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $.

Let $ y = \sin^{-1} x $. Then, we can write:

sin⁡y=x \sin y = x

Differentiating both sides with respect to $ x $, we get:

cos⁡ydydx=1 \cos y \frac{dy}{dx} = 1

Now, we can solve for $ \frac{dy}{dx} $:

dydx=1cos⁡y \frac{dy}{dx} = \frac{1}{\cos y}

But, we want to express $ \frac{dy}{dx} $ in terms of $ x $, not $ y $. To do this, we can use the fact that $ \cos y = \sqrt{1 - \sin^2 y} $.

Substituting this expression into the previous equation, we get:

dydx=11−sin⁡2y \frac{dy}{dx} = \frac{1}{\sqrt{1 - \sin^2 y}}

Now, we can substitute $ \sin y = x $ into this equation:

dydx=11−x2 \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}

Proof of the Derivative

To prove that the derivative of the inverse sine function is indeed $ y' = \frac{1}{\sqrt{1-x^2}} $, we can use the following steps:

  1. Differentiate the sine function: We know that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $.
  2. Use the chain rule: We can use the chain rule to differentiate the inverse sine function:

dydx=ddxsin⁡−1x=1cos⁡y \frac{dy}{dx} = \frac{d}{dx} \sin^{-1} x = \frac{1}{\cos y}

  1. Express $ \cos y $ in terms of $ x $: We can use the fact that $ \cos y = \sqrt{1 - \sin^2 y} $ to express $ \cos y $ in terms of $ x $:

cos⁡y=1−sin⁡2y=1−x2 \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}

  1. Substitute $ \cos y $ into the equation: We can substitute $ \cos y = \sqrt{1 - x^2} $ into the equation:

dydx=1cos⁡y=11−x2 \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}

Conclusion

In this article, we have proved that the derivative of the inverse sine function is $ y' = \frac{1}{\sqrt{1-x^2}} $. We used the chain rule and the fact that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $ to derive this result. This derivative is a fundamental concept in calculus and has many applications in mathematics and physics.

Applications of the Derivative

The derivative of the inverse sine function has many applications in mathematics and physics. Some of these applications include:

  • Trigonometry: The derivative of the inverse sine function is used to find the derivative of trigonometric functions, such as the sine and cosine functions.
  • Calculus: The derivative of the inverse sine function is used to find the derivative of inverse trigonometric functions, such as the inverse sine and inverse cosine functions.
  • Physics: The derivative of the inverse sine function is used to describe the motion of objects in physics, such as the motion of a pendulum.

Example Problems

Here are some example problems that illustrate the use of the derivative of the inverse sine function:

  • Problem 1: Find the derivative of the function $ y = \sin^{-1} x $.
  • Solution: Using the chain rule and the fact that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $, we can find the derivative of the inverse sine function:

dydx=1cos⁡y=11−x2 \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}

  • Problem 2: Find the derivative of the function $ y = \sin^{-1} x $ at the point $ x = 0 $.
  • Solution: Using the derivative of the inverse sine function, we can find the derivative of the function at the point $ x = 0 $:

dydx=11−x2=11−02=1 \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - 0^2}} = 1

Conclusion

Frequently Asked Questions

In this article, we will answer some frequently asked questions about the derivative of the inverse sine function.

Q: What is the derivative of the inverse sine function?

A: The derivative of the inverse sine function is $ y' = \frac{1}{\sqrt{1-x^2}} $.

Q: How do you find the derivative of the inverse sine function?

A: To find the derivative of the inverse sine function, you can use the chain rule and the fact that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $.

Q: What is the domain of the inverse sine function?

A: The domain of the inverse sine function is $ [-1, 1] $.

Q: What is the range of the inverse sine function?

A: The range of the inverse sine function is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.

Q: How do you use the derivative of the inverse sine function in real-world applications?

A: The derivative of the inverse sine function is used in many real-world applications, including:

  • Trigonometry: The derivative of the inverse sine function is used to find the derivative of trigonometric functions, such as the sine and cosine functions.
  • Calculus: The derivative of the inverse sine function is used to find the derivative of inverse trigonometric functions, such as the inverse sine and inverse cosine functions.
  • Physics: The derivative of the inverse sine function is used to describe the motion of objects in physics, such as the motion of a pendulum.

Q: What are some common mistakes to avoid when finding the derivative of the inverse sine function?

A: Some common mistakes to avoid when finding the derivative of the inverse sine function include:

  • Not using the chain rule: The chain rule is essential when finding the derivative of the inverse sine function.
  • Not using the fact that the derivative of the sine function is $ \frac{d}{dx} \sin x = \cos x $: This fact is crucial when finding the derivative of the inverse sine function.
  • Not checking the domain and range of the inverse sine function: The domain and range of the inverse sine function are essential when finding its derivative.

Q: How do you check if the derivative of the inverse sine function is correct?

A: To check if the derivative of the inverse sine function is correct, you can use the following steps:

  • Check the domain and range of the inverse sine function: Make sure that the domain and range of the inverse sine function are correct.
  • Check the derivative of the inverse sine function: Make sure that the derivative of the inverse sine function is correct.
  • Check the derivative of the inverse sine function at specific points: Check the derivative of the inverse sine function at specific points, such as $ x = 0 $.

Q: What are some real-world applications of the derivative of the inverse sine function?

A: Some real-world applications of the derivative of the inverse sine function include:

  • Physics: The derivative of the inverse sine function is used to describe the motion of objects in physics, such as the motion of a pendulum.
  • Engineering: The derivative of the inverse sine function is used in engineering to design and analyze systems that involve trigonometric functions.
  • Computer Science: The derivative of the inverse sine function is used in computer science to develop algorithms that involve trigonometric functions.

Conclusion

In conclusion, the derivative of the inverse sine function is $ y' = \frac{1}{\sqrt{1-x^2}} $. This derivative is a fundamental concept in calculus and has many applications in mathematics and physics. We have answered some frequently asked questions about the derivative of the inverse sine function and provided some real-world applications of this derivative.