/*! 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 39 Evaluate the derivative of the f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 39

Evaluate the derivative of the following functions. $$f(s)=\cot ^{-1}\left(e^{s}\right)$$

Short Answer

Expert verified
Question: Determine the derivative of the function $$f(s) = \cot^{-1}(e^s)$$ with respect to s. Answer: The derivative of the function $$f(s) = \cot^{-1}(e^s)$$ with respect to s is $$\frac{df}{ds} = -\frac{e^s}{(e^{2s} + 1)}$$.

Step by step solution

01

Identify the outer and inner functions

The problem consists of two functions. The outer function is the arccotangent function: $$\cot^{-1}(x)$$ and the inner function is the exponential function: $$e^s$$.
02

Find the derivative of the outer function

First, we will find the derivative of the outer function $$\cot^{-1}(x)$$ with respect to x. Recall the derivative of the arccotangent function: $$\frac{d}{dx} \cot^{-1}(x) = -\frac{1}{x^2 + 1}$$. Here, we can replace x by u: $$\frac{d}{du} \cot^{-1}(u) = -\frac{1}{u^2 + 1}$$.
03

Find the derivative of the inner function

Now, we need to find the derivative of the inner function $$e^s$$ with respect to s. Recall the derivative of the exponential function: $$\frac{d}{ds}e^s = e^s$$.
04

Apply the chain rule to find the derivative of f(s)

Finally, we will apply the chain rule, which tells us that $$\frac{df}{ds} = \frac{d}{du}\left(\cot^{-1}(u)\right) \cdot \frac{du}{ds}$$. By substituting the expressions for the derivatives of outer and inner functions from Steps 2 and 3, we get: $$\frac{df}{ds} = \left(-\frac{1}{u^2 + 1}\right) \cdot e^s$$ Note that the inner function is $$u = e^s$$. We substitute this back into the expression for the derivative to obtain: $$\frac{df}{ds} = \left(-\frac{1}{(e^s)^2 + 1}\right) \cdot e^s$$
05

Final expression for the derivative

Now, we have the final expression for the derivative of the given function: $$\frac{df}{ds} = -\frac{e^s}{(e^{2s} + 1)}$$

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 91影视!

Key Concepts

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

Chain Rule
The Chain Rule is an essential tool in calculus for finding the derivative of a composite function, which is a function made up of two or more functions. It allows us to take the derivative of an outer function while considering how the inner function affects it. Simply put, it helps us differentiate expressions like \(f(g(x))\).
Imagine you have a function \(h(x) = f(g(x))\). The Chain Rule states that the derivative of \(h\) with respect to \(x\) is:
  • First, find the derivative of the outer function \(f\) with respect to its input \(g(x)\), expressed as \(f'(g(x))\).
  • Then, multiply this by the derivative of the inner function \(g(x)\) with respect to \(x\), expressed as \(g'(x)\).
This gives the formula: \( h'(x) = f'(g(x)) \cdot g'(x) \). This technique eases the process of differentiation for complex functions by breaking them into simpler parts.
Arccotangent Function
The arccotangent function, denoted as \(\cot^{-1}(x)\) or \(\text{arccot}(x)\), is the inverse of the cotangent function. It is used to find an angle whose cotangent is \(x\). This function is less common compared to other trigonometric inverses but crucial in calculus.
Its derivative with respect to \(x\) is an important concept. The derivative of \(\cot^{-1}(x)\) is given by:
  • \(\frac{d}{dx} \cot^{-1}(x) = -\frac{1}{x^2 + 1}\).
This derivative is used extensively when differentiating expressions that involve arccotangent, especially when it serves as the outer function of a composite function like in our example \( \cot^{-1}(e^s)\). Understanding how to derive this function helps simplify complex differentiation problems.
Exponential Function
The exponential function is a fundamental mathematical function that appears frequently in calculus and many real-world applications. Represented as \(e^x\), where \(e\) is Euler's number (approximately 2.718), it is noteworthy because its rate of growth is proportional to its current value. This unique property makes it deeply significant in fields like biology, economics, and physics.
When it comes to differentiation, the exponential function has a straightforward derivative:
  • The derivative of \(e^s\) with respect to \(s\) is simply \(e^s\).
This property greatly simplifies calculations involving the derivative because the function remains unchanged. Therefore, when taking a derivative such as \(\cot^{-1}(e^s)\), the exponential nature of one component of the function provides an uncomplicated derivative which pairs well with the Chain Rule, streamlining the solution's process.

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

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. Find the second derivative \(\frac{d^{2} y}{d t^{2}}\) b. Verify that \(\frac{d^{2} y}{d t^{2}}=-\frac{k}{m} y\)

\- 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 point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. $$3 x^{3}+7 y^{3}=10 y; \left(x_{0}, y_{0}\right)=(1,1)$$ (Graph cant copy)

Suppose \(f\) is differentiable for all real numbers with \(f(0)=-3, f(1)=3, f^{\prime}(0)=3,\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=\sin (\pi f(x)) .\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}(1)\)

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x+y=\cos y$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation
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.