Find The Derivative Of The Function:$\[ Y = \cot^2(\sin(\theta)) \\]Calculate $\[ Y' = \square \\]

Introduction

In this article, we will delve into the world of calculus and explore the concept of finding the derivative of a trigonometric function. Specifically, we will focus on the function $y = \cot^2(\sin(\theta))$ and calculate its derivative. The derivative of a function represents the rate of change of the function with respect to its input variable. In this case, we will use the chain rule and other trigonometric identities to find the derivative of the given function.

Understanding the Function

Before we dive into the calculation, let's take a closer look at the function $y = \cot^2(\sin(\theta))$. This function involves the composition of three trigonometric functions: the sine function, the cotangent function, and the square function. The sine function takes an angle $\theta$ as input and returns a value between -1 and 1. The cotangent function takes the ratio of the adjacent side to the opposite side of a right triangle as input and returns a value. The square function takes a value as input and returns its square.

Applying the Chain Rule

To find the derivative of the function $y = \cot^2(\sin(\theta))$, we will use the chain rule. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f'(g(x)) \cdot g'(x)$. In this case, we can identify the outer function as $f(u) = u^2$ and the inner function as $g(\theta) = \cot^2(\sin(\theta))$.

Finding the Derivative of the Outer Function

The derivative of the outer function $f(u) = u^2$ is given by $f'(u) = 2u$. We can substitute $u = \cot^2(\sin(\theta))$ into this expression to get $f'(\cot^2(\sin(\theta))) = 2\cot^2(\sin(\theta))$.

Finding the Derivative of the Inner Function

The derivative of the inner function $g(\theta) = \cot^2(\sin(\theta))$ is not as straightforward. We will need to use the chain rule again to find the derivative of the inner function. Let's identify the outer function as $h(v) = \cot^2(v)$ and the inner function as $k(\theta) = \sin(\theta)$.

Finding the Derivative of the Outer Function of the Inner Function

The derivative of the outer function $h(v) = \cot^2(v)$ is given by $h'(v) = -2\cot(v)\csc^2(v)$. We can substitute $v = \sin(\theta)$ into this expression to get $h'(\sin(\theta)) = -2\cot(\sin(\theta))\csc^2(\sin(\theta))$.

Finding the Derivative of the Inner Function of the Inner Function

The derivative of the inner function $k(\theta) = \sin(\theta)$ is given by $k'(\theta) = \cos(\theta)$.

Applying the Chain Rule Again

Now that we have found the derivatives of the outer and inner functions, we can apply the chain rule again to find the derivative of the inner function $g(\theta) = \cot^2(\sin(\theta))$. We have $g'(\theta) = h'(\sin(\theta)) \cdot k'(\theta) = -2\cot(\sin(\theta))\csc^2(\sin(\theta)) \cdot \cos(\theta)$.

Finding the Derivative of the Original Function

Now that we have found the derivative of the inner function, we can substitute it into the expression for the derivative of the outer function to get the final derivative of the original function. We have $y' = f'(\cot^2(\sin(\theta))) \cdot g'(\theta) = 2\cot^2(\sin(\theta)) \cdot (-2\cot(\sin(\theta))\csc^2(\sin(\theta)) \cdot \cos(\theta))$.

Simplifying the Derivative

We can simplify the derivative by combining like terms. We have $y' = -4\cot^3(\sin(\theta))\csc^2(\sin(\theta)) \cdot \cos(\theta)$.

Conclusion

In this article, we have found the derivative of the function $y = \cot^2(\sin(\theta))$ using the chain rule and other trigonometric identities. The derivative of the function is given by $y' = -4\cot^3(\sin(\theta))\csc^2(\sin(\theta)) \cdot \cos(\theta)$. This result demonstrates the power of the chain rule in finding the derivatives of composite functions.

Final Answer

The final answer is: $\boxed{-4\cot^3(\sin(\theta))\csc^2(\sin(\theta)) \cdot \cos(\theta)}$

Introduction

In our previous article, we explored the concept of finding the derivative of a trigonometric function, specifically the function $y = \cot^2(\sin(\theta))$. We used the chain rule and other trigonometric identities to find the derivative of the function. In this article, we will answer some common questions related to finding the derivative of a trigonometric function.

Q: What is the chain rule?

A: The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. A composite function is a function that is composed of two or more functions. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of the function is given by $f'(g(x)) \cdot g'(x)$.

Q: How do I apply the chain rule?

A: To apply the chain rule, you need to identify the outer function and the inner function. The outer function is the function that is being composed, and the inner function is the function that is being composed with. Once you have identified the outer and inner functions, you can find the derivative of the outer function and the derivative of the inner function. Then, you can multiply the derivatives together to get the final derivative.

Q: What are some common trigonometric identities that I need to know?

A: There are several common trigonometric identities that you need to know when finding the derivative of a trigonometric function. Some of the most common identities include:

• $\sin^2(x) + \cos^2(x) = 1$
• $\tan(x) = \frac{\sin(x)}{\cos(x)}$
• $\cot(x) = \frac{\cos(x)}{\sin(x)}$
• $\sec(x) = \frac{1}{\cos(x)}$
• $\csc(x) = \frac{1}{\sin(x)}$

Q: How do I find the derivative of a trigonometric function that involves a square?

A: To find the derivative of a trigonometric function that involves a square, you need to use the chain rule and the power rule. The power rule states that if we have a function of the form $f(x^n)$, then the derivative of the function is given by $nf(x^{n-1})$. In the case of a trigonometric function that involves a square, you can use the chain rule to find the derivative of the outer function and the power rule to find the derivative of the inner function.

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

A: There are several common mistakes to avoid when finding the derivative of a trigonometric function. Some of the most common mistakes include:

• Forgetting to use the chain rule
• Forgetting to use the power rule
• Not simplifying the derivative
• Not checking the domain of the function

Q: How do I check the domain of a trigonometric function?

A: To check the domain of a trigonometric function, you need to make sure that the input values are within the range of the function. For example, the sine function is defined for all real numbers, but the cosine function is only defined for angles between 0 and 360 degrees. You need to check the domain of the function before finding the derivative.

Conclusion

In this article, we have answered some common questions related to finding the derivative of a trigonometric function. We have discussed the chain rule, common trigonometric identities, and common mistakes to avoid. We have also provided some tips for checking the domain of a trigonometric function. By following these tips and avoiding common mistakes, you can find the derivative of a trigonometric function with confidence.

Final Answer

The final answer is: There is no final numerical answer to this problem. The article is a Q&A guide to finding the derivative of a trigonometric function.