/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 In Exercises \(41-58,\) find \(d... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercises \(41-58,\) find \(d y / d t\) $$y=(1+\cot (t / 2))^{-2}$$

Short Answer

Expert verified
\( \frac{dy}{dt} = (1 + \cot(t/2))^{-3} \csc^2(t/2) \).

Step by step solution

01

Identify the Problem

We need to find the derivative of the function \( y = (1 + \cot(t/2))^{-2} \) with respect to \( t \). This requires using chain rule and trigonometric identities.
02

Apply the Chain Rule

Let \( u = 1 + \cot(t/2) \). Then \( y = u^{-2} \). By the chain rule, \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \). First, differentiate \( y = u^{-2} \) to get \( \frac{dy}{du} = -2u^{-3} \).
03

Differentiate the Inner Function

Now, differentiate \( u = 1 + \cot(t/2) \) with respect to \( t \). Use the derivative of cotangent: \( \frac{d}{dt} [\cot(t/2)] = -\csc^2(t/2) \cdot \frac{1}{2} \). Therefore, \( \frac{du}{dt} = -\frac{1}{2} \csc^2(t/2) \).
04

Combine Derivatives

Substitute \( \frac{dy}{du} = -2u^{-3} \) and \( \frac{du}{dt} = -\frac{1}{2} \csc^2(t/2) \) into the chain rule formula: \[ \frac{dy}{dt} = (-2u^{-3}) \cdot \left(-\frac{1}{2} \csc^2(t/2)\right), \]which simplifies to \[ \frac{dy}{dt} = u^{-3} \csc^2(t/2). \]
05

Substitute Back for \( u \)

Recall \( u = 1 + \cot(t/2) \). Substitute back to express \( \frac{dy}{dt} \) in terms of \( t \): \[ \frac{dy}{dt} = (1 + \cot(t/2))^{-3} \csc^2(t/2). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule in Differentiation
The chain rule is a foundational concept in calculus that helps us differentiate composite functions. A composite function is one that can be represented as a function within another function, like peeling layers of an onion. To apply the chain rule, you need to first identify the "outer" function and the "inner" function. For instance, imagine the function \( y = (1 + \cot(t/2))^{-2} \), where the outer function is \( u^{-2} \) and the inner function is \( u = 1 + \cot(t/2) \).
\[
\text{Chain rule formula:} \, \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}
\]
To find \( \frac{dy}{di} \), we first differentiate \( u^{-2} \) with respect to \( u \), giving us \( -2u^{-3} \). Next, we find \( \frac{du}{dt} \) by differentiating the inner function \( 1 + \cot(t/2) \), which involves more specifics we'll explore further. Finally, multiply these derivatives to obtain the overall result. This method is vital, especially in calculus classes, as it simplifies the process of finding derivatives of complex-looking functions.
Understanding Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables where both sides are defined. Understanding these identities is crucial in calculus when differentiating functions involving trigonometric terms. Common identities used include the Pythagorean identity and identities related to complementary angles.
For instance, in this exercise, we're dealing with the cotangent function, \( \cot(\theta) \), which is defined as the reciprocal of the tangent function:
\[ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)} \]
Additionally, the derivative of \( \cot(\theta) \) with respect to \( \theta \) is \( -\csc^2(\theta) \). This identity is key when working out derivatives of functions like \( 1 + \cot(t/2) \), as it helps in calculating \( \frac{du}{dt} \). Trigonometric identities simplify otherwise complex differentiation tasks, providing a toolkit for tackling intricate problems.
Cotangent Function in Calculus
The cotangent function, written as \( \cot(\theta) \), is one of the basic trigonometric functions. In a right-angled triangle, it is the ratio of the length of the adjacent side to the opposite side. However, in calculus, its role extends beyond simple triangles. The cotangent function becomes instrumental when analyzing periodic patterns and waves, especially in contexts involving angular motion.
When differentiating expressions with cotangent functions, such as \( 1 + \cot(t/2) \), knowing how to differentiate \( \cot \) itself is vital. The derivative is \( \frac{d}{dt} [\cot(t/2)] = -\csc^2(t/2) \cdot \frac{1}{2} \), as the angle is modified by dividing by 2. This rule arises from the chain rule combined with the derivative of the cotangent, which is \( -\csc^2(\theta) \), another trigonometric identity.
Understanding these derivatives and identities allows you to navigate complex trigonometric equations with greater ease, enhancing both computation and conceptual comprehension in calculus.

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Most popular questions from this chapter

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