What Is The Value Of D D X Csc ⁡ ( X \frac{d}{d X} \csc (x D X D ​ Csc ( X ] At X = Π 6 X=\frac{\pi}{6} X = 6 Π ​ ?Choose One Answer:A. − 2 3 -2 \sqrt{3} − 2 3 ​ B. 1 2 \frac{1}{2} 2 1 ​ C. − 3 -\sqrt{3} − 3 ​ D. 2

What is the Value of $\frac{d}{d x} \csc (x)$ at $x=\frac{\pi}{6}$?

In this article, we will explore the concept of finding the derivative of the cosecant function and evaluate it at a specific point. The cosecant function is the reciprocal of the sine function, and its derivative is an essential concept in calculus. We will use the chain rule and the definition of the derivative to find the derivative of the cosecant function and then evaluate it at $x=\frac{\pi}{6}$.

The cosecant function is defined as:

$$\csc (x) = \frac{1}{\sin (x)}$$

This function is the reciprocal of the sine function, and it is periodic with a period of $2\pi$. The cosecant function has a vertical asymptote at $x=0$ and $x=\pi$.

To find the derivative of the cosecant function, we will use the chain rule and the definition of the derivative. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the composite function is given by:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$$

In this case, we have:

$$\csc (x) = \frac{1}{\sin (x)}$$

We can rewrite this as:

$$\csc (x) = f(\sin (x))$$

where $f(u) = \frac{1}{u}$. The derivative of $f(u)$ is:

$$f'(u) = -\frac{1}{u^2}$$

Using the chain rule, we get:

$$\frac{d}{dx} \csc (x) = f'(\sin (x)) \cdot \sin'(x)$$

$$\frac{d}{dx} \csc (x) = -\frac{1}{\sin^2 (x)} \cdot \cos (x)$$

$$\frac{d}{dx} \csc (x) = -\frac{\cos (x)}{\sin^2 (x)}$$

$$

Now that we have found the derivative of the cosecant function, we can evaluate it at $x=\frac{\pi}{6}$. We know that:

$$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$$

$$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

Substituting these values into the derivative, we get:

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2}$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\frac{\sqrt{3}}{2}}{\frac{1}{4}}$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \cdot 4$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -2\sqrt{3}$$

Therefore, the value of $\frac{d}{dx} \csc (x)$ at $x=\frac{\pi}{6}$ is $-2\sqrt{3}$.

In this article, we found the derivative of the cosecant function using the chain rule and the definition of the derivative. We then evaluated the derivative at $x=\frac{\pi}{6}$ and found that the value of $\frac{d}{dx} \csc (x)$ at $x=\frac{\pi}{6}$ is $-2\sqrt{3}$. This result is consistent with the options provided in the discussion category.

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

In our previous article, we explored the concept of finding the derivative of the cosecant function and evaluated it at a specific point. In this article, we will answer some frequently asked questions related to the derivative of the cosecant function.

Q: What is the derivative of the cosecant function?

A: The derivative of the cosecant function is given by:

$$\frac{d}{dx} \csc (x) = -\frac{\cos (x)}{\sin^2 (x)}$$

Q: How do I find the derivative of the cosecant function?

A: To find the derivative of the cosecant function, you can use the chain rule and the definition of the derivative. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the composite function is given by:

$$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$$

In this case, we have:

$$\csc (x) = \frac{1}{\sin (x)}$$

We can rewrite this as:

$$\csc (x) = f(\sin (x))$$

where $f(u) = \frac{1}{u}$. The derivative of $f(u)$ is:

$$f'(u) = -\frac{1}{u^2}$$

Using the chain rule, we get:

$$\frac{d}{dx} \csc (x) = f'(\sin (x)) \cdot \sin'(x)$$

$$\frac{d}{dx} \csc (x) = -\frac{1}{\sin^2 (x)} \cdot \cos (x)$$

Q: What is the value of the derivative of the cosecant function at $x=\frac{\pi}{6}$?

A: The value of the derivative of the cosecant function at $x=\frac{\pi}{6}$ is $-2\sqrt{3}$.

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

A: To evaluate the derivative of the cosecant function at a specific point, you can substitute the value of the point into the derivative. For example, to evaluate the derivative of the cosecant function at $x=\frac{\pi}{6}$, you can substitute $\frac{\pi}{6}$ into the derivative:

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2}$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\frac{\sqrt{3}}{2}}{\frac{1}{4}}$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \cdot 4$$

$$\frac{d}{dx} \csc \left(\frac{\pi}{6}\right) = -2\sqrt{3}$$

Q: What are some common applications of the derivative of the cosecant function?

A: The derivative of the cosecant function has many applications in mathematics and physics. Some common applications include:

• Finding the rate of change of the cosecant function
• Evaluating the derivative of the cosecant function at specific points
• Using the derivative of the cosecant function to solve optimization problems
• Using the derivative of the cosecant function to model real-world phenomena

In this article, we answered some frequently asked questions related to the derivative of the cosecant function. We hope that this article has been helpful in clarifying any confusion you may have had about the derivative of the cosecant function. If you have any further questions, please don't hesitate to ask.

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

The content of this article is for educational purposes only and is not intended to be used as a substitute for professional advice or guidance.