/*! 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 23 Evaluate the derivatives of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the derivatives of the following functions. $$f(y)=\cot ^{-1}\left(1 /\left(y^{2}+1\right)\right)$$

Short Answer

Expert verified
Answer: The derivative of the function is $$f'(y) = \frac{2y}{(y^2 + 1)^3 + (y^2 + 1)}$$.

Step by step solution

01

Identify the outer and inner functions

In our given function $$f(y) = \cot^{-1}\left(\frac{1}{y^2 + 1}\right)$$, the outer function is the inverse cotangent function denoted as $$\cot^{-1}(u)$$ and the inner function is $$u = \frac{1}{y^2 + 1}$$. We will find the derivative of both functions separately.
02

Derivative of the outer function (inverse cotangent)

Recall that the derivative of the inverse cotangent function is given by: $$\frac{d\cot^{-1}(u)}{du} = -\frac{1}{u^2+1}$$. We will apply this to the outer function.
03

Derivative of the inner function

Now, we need to find the derivative of the inner function $$u = \frac{1}{y^2 + 1}$$. To do this, we can use the Power Rule: $$\frac{d}{dy}\left(\frac{1}{y^2 + 1}\right) = \frac{-2y}{(y^2+1)^2}$$.
04

Apply the chain rule to find the derivative of the function

Finally, we will now apply the chain rule to find the derivative of the given function $$f(y) = \cot^{-1}\left(\frac{1}{y^2 + 1}\right)$$. By using the chain rule, we get: $$\frac{df(y)}{dy} = \frac{d(\cot^{-1}(u))}{du} \times \frac{du}{dy}$$. Now, substitute the results from Step 2 and Step 3: #tag_end_title# $$\frac{df(y)}{dy} = \left(-\frac{1}{u^2+1}\right)\left(\frac{-2y}{(y^2+1)^2}\right)$$. Replace $$u$$ with its corresponding expression: $$\frac{df(y)}{dy} = -\frac{(-2y)}{((y^2+1)^2)(\left(\frac{1}{y^2+1}\right)^2+1)}$$.
05

Simplify the derivative expression

After substituting and simplifying the expression to its final form, we get: $$\frac{df(y)}{dy} = \frac{2y}{(y^2 + 1)^3 + (y^2 + 1)}$$. The final answer for the derivative of the function is: $$f'(y) = \frac{2y}{(y^2 + 1)^3 + (y^2 + 1)}$$.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse operations of the trigonometric functions. They allow us to find the angle that corresponds to a given trigonometric ratio. There are six primary inverse trigonometric functions, corresponding to the six trigonometric functions: inverse sine, inverse cosine, inverse tangent, inverse cotangent, inverse secant, and inverse cosecant.

Each of these has a specific domain and range that ensure the function has a proper inverse. For example, the inverse sine function, noted as \( \sin^{-1}(x) \), spans the range \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). Similarly, the inverse cotangent, \( \cot^{-1}(x) \), ranges between 0 and \( \pi \).

These inverse operations are crucial in calculus when we need to evaluate problems involving trigonometric expressions that must return specific angle measures. Knowing how to differentiate them is useful in many physics and engineering applications, making them a key tool in mathematical problem-solving.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When dealing with a function composed of two or more functions, the chain rule helps us differentiate the outer function in terms of the inner function and then multiply this result by the derivative of the inner function.

In simpler terms, if we have a composite function \( f(g(x)) \), the derivative \( \frac{d}{dx}[f(g(x))] \) is \( f'(g(x)) \cdot g'(x) \). This rule is very powerful and frequently employed in calculus, particularly when working with complex expressions that are not straightforward to differentiate individually.

Applying the chain rule requires recognizing the different layers or functions within an expression. Once these are identified, one simply needs to differentiate each layer separately, as seen in the original problem where \( f(y) = \cot^{-1}\left(\frac{1}{y^2 + 1}\right) \). Here, the inverse cotangent is the outer function, and \( \frac{1}{y^2 + 1} \) is the inner function.
Inverse Cotangent Function
The inverse cotangent function, written as \( \cot^{-1}(x) \), is used to find the angle whose cotangent is the value \( x \). This function is less common in trigonometry but still has significant applications in calculus and complex analysis.

The derivative of the inverse cotangent function is crucial for solving calculus problems involving arc cotangent expressions, and it is always given by \( \frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{x^2 + 1} \). This result is derived from the relationship between the trigonometric and inverse trigonometric functions and shows how the slope of the inverse cotangent function behaves for different values of \( x \).

When solving problems like the exercise given, understanding and applying the derivative of \( \cot^{-1}(x) \) allows us to chain together derivatives effectively using the chain rule. In practice, this derivative helps in various applications, such as integration, where knowing these relationships can significantly simplify the steps involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(1+\frac{1}{x}\right)^{x}$$

Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no points of intersection. It follows that for some real number \(1

A store manager estimates that the demand for an energy drink decreases with increasing price according to the function \(d(p)=\frac{100}{p^{2}+1},\) which means that at price \(p\) (in dollars), \(d(p)\) units can be sold. The revenue generated at price \(p\) is \(R(p)=p \cdot d(p)\) (price multiplied by number of units). a. Find and graph the revenue function. b. Find and graph the marginal revenue \(R^{\prime}(p)\). c. From the graphs of \(R\) and \(R^{\prime}\), estimate the price that should be charged to maximize the revenue.

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(1+x^{2}\right)^{\sin x}$$

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.