Find The Derivative Of The Function:$y = \sqrt{1-x^2} - X \cos^{-1}(x$\]

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Introduction

In calculus, the derivative of a function is a measure of how the function changes as its input changes. It is a fundamental concept in mathematics and has numerous applications in various fields, including physics, engineering, and economics. In this article, we will focus on finding the derivative of a complex function, specifically the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x).

Understanding the Function

Before we dive into finding the derivative, let's break down the function and understand its components. The function consists of two parts: 1−x2\sqrt{1-x^2} and −xcos⁡−1(x)-x \cos^{-1}(x). The first part is a square root function, while the second part involves the inverse cosine function.

Square Root Function

The square root function 1−x2\sqrt{1-x^2} is defined as the positive square root of 1−x21-x^2. This function is also known as the absolute value function, and it is defined for all real numbers xx such that −1≤x≤1-1 \leq x \leq 1.

Inverse Cosine Function

The inverse cosine function cos⁡−1(x)\cos^{-1}(x) is defined as the angle whose cosine is xx. This function is also known as the arccosine function, and it is defined for all real numbers xx such that −1≤x≤1-1 \leq x \leq 1.

Finding the Derivative

To find the derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x), we will use the chain rule and the product rule of differentiation.

Step 1: Differentiate the Square Root Function

To differentiate the square root function 1−x2\sqrt{1-x^2}, we will use the chain rule. The derivative of the square root function is given by:

ddx1−x2=12(1−x2)−12(−2x)\frac{d}{dx} \sqrt{1-x^2} = \frac{1}{2} (1-x^2)^{-\frac{1}{2}} (-2x)

Simplifying the expression, we get:

ddx1−x2=−x1−x2\frac{d}{dx} \sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}}

Step 2: Differentiate the Inverse Cosine Function

To differentiate the inverse cosine function cos⁡−1(x)\cos^{-1}(x), we will use the chain rule. The derivative of the inverse cosine function is given by:

ddxcos⁡−1(x)=−11−x2\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}

Step 3: Apply the Product Rule

To find the derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x), we will apply the product rule of differentiation. The product rule states that if y=u(x)v(x)y = u(x)v(x), then y′=u′(x)v(x)+u(x)v′(x)y' = u'(x)v(x) + u(x)v'(x).

Applying the product rule, we get:

y′=ddx1−x2−xddxcos⁡−1(x)y' = \frac{d}{dx} \sqrt{1-x^2} - x \frac{d}{dx} \cos^{-1}(x)

Substituting the derivatives of the square root function and the inverse cosine function, we get:

y′=−x1−x2−x(−11−x2)y' = -\frac{x}{\sqrt{1-x^2}} - x \left(-\frac{1}{\sqrt{1-x^2}}\right)

Simplifying the expression, we get:

y′=−x1−x2+x1−x2y' = -\frac{x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}

y′=0y' = 0

Conclusion

In this article, we found the derivative of the complex function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x). We used the chain rule and the product rule of differentiation to find the derivative of the function. The final answer is y′=0y' = 0.

Limitations

The function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x) is defined for all real numbers xx such that −1≤x≤1-1 \leq x \leq 1. However, the derivative of the function is only defined for all real numbers xx such that −1<x<1-1 < x < 1.

Future Work

In future work, we can explore other complex functions and find their derivatives using the chain rule and the product rule of differentiation. We can also investigate the properties of the derivatives of these functions and their applications in various fields.

References

  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Advanced Calculus, 2nd edition, Michael Spivak

Appendix

The following is a list of the formulas and theorems used in this article:

  • Chain rule: ddxf(g(x))=f′(g(x))⋅g′(x)\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)
  • Product rule: ddxu(x)v(x)=u′(x)v(x)+u(x)v′(x)\frac{d}{dx} u(x)v(x) = u'(x)v(x) + u(x)v'(x)
  • Derivative of the square root function: ddx1−x2=−x1−x2\frac{d}{dx} \sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}}
  • Derivative of the inverse cosine function: ddxcos⁡−1(x)=−11−x2\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}
    Derivative of a Complex Function: A Q&A Guide =====================================================

Introduction

In our previous article, we found the derivative of the complex function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x). However, we understand that some readers may still have questions about the derivative of this function. In this article, we will address some of the most frequently asked questions about the derivative of this function.

Q&A

Q: What is the derivative of the square root function?

A: The derivative of the square root function 1−x2\sqrt{1-x^2} is given by:

ddx1−x2=−x1−x2\frac{d}{dx} \sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}}

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

A: The derivative of the inverse cosine function cos⁡−1(x)\cos^{-1}(x) is given by:

ddxcos⁡−1(x)=−11−x2\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}

Q: How do I apply the product rule to find the derivative of the function?

A: To find the derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x), you can apply the product rule of differentiation. The product rule states that if y=u(x)v(x)y = u(x)v(x), then y′=u′(x)v(x)+u(x)v′(x)y' = u'(x)v(x) + u(x)v'(x).

Q: What is the final answer to the derivative of the function?

A: The final answer to the derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x) is y′=0y' = 0.

Q: What are the limitations of the derivative of the function?

A: The derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x) is only defined for all real numbers xx such that −1<x<1-1 < x < 1.

Q: Can I use the chain rule to find the derivative of the function?

A: Yes, you can use the chain rule to find the derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x). The chain rule states that if y=f(g(x))y = f(g(x)), then y′=f′(g(x))⋅g′(x)y' = f'(g(x)) \cdot g'(x).

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

A: The derivative of the function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x) has numerous real-world applications in various fields, including physics, engineering, and economics.

Conclusion

In this article, we addressed some of the most frequently asked questions about the derivative of the complex function y=1−x2−xcos⁡−1(x)y = \sqrt{1-x^2} - x \cos^{-1}(x). We hope that this article has provided you with a better understanding of the derivative of this function and its applications.

References

  • [1] Calculus, 3rd edition, Michael Spivak
  • [2] Calculus, 2nd edition, James Stewart
  • [3] Advanced Calculus, 2nd edition, Michael Spivak

Appendix

The following is a list of the formulas and theorems used in this article:

  • Chain rule: ddxf(g(x))=f′(g(x))⋅g′(x)\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)
  • Product rule: ddxu(x)v(x)=u′(x)v(x)+u(x)v′(x)\frac{d}{dx} u(x)v(x) = u'(x)v(x) + u(x)v'(x)
  • Derivative of the square root function: ddx1−x2=−x1−x2\frac{d}{dx} \sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}}
  • Derivative of the inverse cosine function: ddxcos⁡−1(x)=−11−x2\frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}

Glossary

  • Derivative: A measure of how a function changes as its input changes.
  • Chain rule: A rule of differentiation that states that if y=f(g(x))y = f(g(x)), then y′=f′(g(x))⋅g′(x)y' = f'(g(x)) \cdot g'(x).
  • Product rule: A rule of differentiation that states that if y=u(x)v(x)y = u(x)v(x), then y′=u′(x)v(x)+u(x)v′(x)y' = u'(x)v(x) + u(x)v'(x).
  • Square root function: A function that returns the positive square root of a number.
  • Inverse cosine function: A function that returns the angle whose cosine is a given number.