The Derivative of the Inverse Sine Function

Oct 13, 2025 by pynkocean.com 1 views

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 evaluate the derivative of the inverse sine function at a specific point, $x = \frac{1}{2}$ .

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

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}

This is because the derivative of the sine function is the cosine function, and the derivative of the inverse sine function is the reciprocal of the square root of $1-x^2$ .

**Evaluating the Derivative at $x = \frac{1}{2}$

Now, we need to evaluate the derivative of the inverse sine function at $x = \frac{1}{2}$ . We can plug in this value into the formula we derived earlier:

\left.\frac{dy}{dx}\right|_{x=\frac{1}{2}} = \frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}}

Simplifying this expression, we get:

\left.\frac{dy}{dx}\right|_{x=\frac{1}{2}} = \frac{1}{\sqrt{1-\frac{1}{4}}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}

Simplifying the Expression

We can simplify this expression further by rationalizing the denominator. To do this, we multiply the numerator and denominator by $\sqrt{3}$ :

\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Conclusion

In this article, we evaluated the derivative of the inverse sine function at a specific point, $x = \frac{1}{2}$ . We used the chain rule and the fact that the derivative of the sine function is the cosine function to derive the formula for the derivative of the inverse sine function. We then plugged in the value $x = \frac{1}{2}$ into this formula and simplified the resulting expression.

The Final Answer

The final answer is $\boxed{\frac{2\sqrt{3}}{3}}$ .

Discussion

The derivative of the inverse sine function is an important concept in mathematics, particularly in calculus. It is used to find the rate of change of the inverse sine function with respect to the input variable. In this article, we evaluated the derivative of the inverse sine function at a specific point, $x = \frac{1}{2}$ . We used the chain rule and the fact that the derivative of the sine function is the cosine function to derive the formula for the derivative of the inverse sine function.

References

[1] "Calculus" by Michael Spivak
[2] "Calculus: Early Transcendentals" by James Stewart
[3] "Mathematics for the Nonmathematician" by Morris Kline
Evaluating the Derivative of the Inverse Sine Function at a Specific Point: Q&A ====================================================================================

Introduction

In our previous article, we evaluated the derivative of the inverse sine function at a specific point, $x = \frac{1}{2}$ . We used the chain rule and the fact that the derivative of the sine function is the cosine function to derive the formula for the derivative of the inverse sine function. In this article, we will answer some common questions related to this topic.

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

A: The derivative of the inverse sine function is given by the formula:

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}

Q: How do I evaluate the derivative of the inverse sine function at a specific point?

A: To evaluate the derivative of the inverse sine function at a specific point, you can plug in the value of $x$ into the formula:

\left.\frac{dy}{dx}\right|_{x=a} = \frac{1}{\sqrt{1-a^2}}

Q: What is the value of the derivative of the inverse sine function at $x = \frac{1}{2}$ ?

A: The value of the derivative of the inverse sine function at $x = \frac{1}{2}$ is:

\left.\frac{dy}{dx}\right|_{x=\frac{1}{2}} = \frac{2}{\sqrt{3}}

Q: How do I simplify the expression for the derivative of the inverse sine function?

A: To simplify the expression for the derivative of the inverse sine function, you can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$ :

\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

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

A: The derivative of the inverse sine function is an important concept in mathematics, particularly in calculus. It is used to find the rate of change of the inverse sine function with respect to the input variable.

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

A: The derivative of the inverse sine function can be used in a variety of real-world applications, such as:

Finding the rate of change of the inverse sine function with respect to the input variable
Modeling the behavior of physical systems that involve the inverse sine function
Solving optimization problems that involve the inverse sine function

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

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

Failing to rationalize the denominator
Failing to simplify the expression
Using the wrong formula for the derivative of the inverse sine function

Conclusion

In this article, we answered some common questions related to the derivative of the inverse sine function. We provided formulas and examples to help illustrate the concepts, and we discussed some common mistakes to avoid when evaluating the derivative of the inverse sine function.

References

[1] "Calculus" by Michael Spivak
[2] "Calculus: Early Transcendentals" by James Stewart
[3] "Mathematics for the Nonmathematician" by Morris Kline