/*! 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 68 In Exercises \(49-70\) , find th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x $$

Short Answer

Expert verified
The derivative \( \frac{dy}{dx} \) is \( \frac{x^2 - 1}{x^2 + 1} \).

Step by step solution

01

Recall the Derivative Formulas

The derivative of the inverse cotangent function, \( \cot^{-1}(u) \), is \( \frac{-1}{1+u^2} \frac{du}{dx} \). Similarly, the derivative of the inverse tangent function, \( \tan^{-1}(u) \), is \( \frac{1}{1+u^2} \frac{du}{dx} \). We will apply these formulas to find \( \frac{dy}{dx} \).
02

Find the Derivative of \( \cot^{-1}\left(\frac{1}{x}\right) \)

Let \( u = \frac{1}{x} \). Then \( \frac{du}{dx} = -\frac{1}{x^2} \). Using the derivative formula for \( \cot^{-1}(u) \), we have:\[\frac{d}{dx} \left( \cot^{-1}\left(\frac{1}{x}\right) \right) = \frac{-1}{1+\left(\frac{1}{x}\right)^2} \left(-\frac{1}{x^2}\right) = \frac{1}{1 + \frac{1}{x^2}} \cdot \frac{1}{x^2}\]Simplifying, we get:\[\frac{x^2}{x^2 + 1}\]
03

Find the Derivative of \( \tan^{-1}(x) \)

Use the derivative formula for the inverse tangent function:\[\frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1+x^2}\]
04

Combine the Results

The derivative of the given function \( y = \cot^{-1}\left(\frac{1}{x}\right) - \tan^{-1}(x) \) is the difference of the two derivatives computed:\[\frac{dy}{dx} = \frac{x^2}{x^2 + 1} - \frac{1}{1+x^2}\]Simplifying:\[\frac{x^2}{x^2 + 1} - \frac{1}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}\]
05

Conclusion

Therefore, the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{x^2 - 1}{x^2 + 1} \).

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 standard trigonometric functions. They help us solve equations where the angle is the unknown we need to find. The inverse cotangent, denoted as \( \cot^{-1}(x) \), and the inverse tangent, denoted as \( \tan^{-1}(x) \), are examples of these functions.
Understanding their properties and derivatives is essential for solving calculus problems involving inverse trigonometric functions.
  • \( \cot^{-1}(x) \) gives us the angle whose cotangent is \( x \).
  • \( \tan^{-1}(x) \) gives us the angle whose tangent is \( x \).
These functions have specific derivative formulas which play a crucial role when differentiating expressions involving them.
Derivative Formulas
Derivative formulas are like handy rules that help us differentiate functions easily. When it comes to inverse trigonometric functions, remembering their derivative formulas is vital.For the inverse cotangent function \( \cot^{-1}(u) \), the derivative is:\[\frac{d}{dx} \left( \cot^{-1}(u) \right) = \frac{-1}{1+u^2} \cdot \frac{du}{dx}\]Similarly, the derivative for the inverse tangent function \( \tan^{-1}(u) \) is:\[\frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1+u^2} \cdot \frac{du}{dx}\]
  • These formulas are helpful as they allow substitution and further simplification when taking derivatives.
  • In the original problem, identifying \( u \) and computing \( \frac{du}{dx} \) were crucial steps for applying these formulas.
Learning these formulas helps save time when tackling calculus problems.
Calculus Problem Solving
Solving calculus problems often involves applying derivatives and simplifications to reach the final answer efficiently. Here are a few key steps from the original problem solution:1. **Understand the Problem:** We need to find the derivative of a function involving inverse trigonometric functions.2. **Apply Derivative Formulas:** Use the derivative formulas for \( \cot^{-1}\left(\frac{1}{x}\right) \) and \( \tan^{-1}(x) \). We find this by setting substitution for variables and differentiating accordingly.3. **Simplify the Result:** After applying the derivatives, the expression might look complex. Simplifying it gives a more straightforward and more comprehensible result.For the original function, the derivative got simplified from:\[\frac{dy}{dx} = \frac{x^2}{x^2 + 1} - \frac{1}{1+x^2}\]to:\[\frac{x^2 - 1}{x^2 + 1}\]
  • This step is crucial as it not only finds the derivative but optimizes its form, which might be necessary for further computations or analyses.
By following these problem-solving steps effectively, understanding and solving calculus problems becomes much easier. Remember, practice is key to mastering these methods!

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

When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logarithms, as shown here. $$ \begin{array}{rlrl}{\sinh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {-\infty < x < \infty} \\ {\cosh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {x \geq 1} \\ {\tanh ^{-1} x} & {=\frac{1}{2} \ln \frac{1+x}{1-x},} & {|x| < 1} \\ {\operatorname{sech}^{-1} x} & {=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right),} & {0 < x \leq 1} \\\ {\operatorname{csch}^{-1} x} & {=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right),} & {x \neq 0} \\\ {\operatorname{coth}^{-1} x} & {=\frac{1}{2} \ln \frac{x+1}{x-1},} & {|x| > 1}\end{array} $$ Use the formulas given above to express the numbers in Exercises \(61-66\) in terms of natural logarithms. $$ \tanh ^{-1}(-1 / 2) $$

A minimal surface Find the area of the surface swept out by revolving about the \(x\) -axis the curve \(y=4 \cosh (x / 4)\) \(-\ln 16 \leq x \leq \ln 81 .\) Graph cannot copy It can be shown that, of all continuously differentiable curves joining points \(A\) and \(B\) in the figure, the curve \(y=4 \cosh (x / 4)\) generates the surface of least area. If you made a rigid wire frame of the end-circles through \(A\) and \(B\) and dipped them in a soap-film solution, the surface spanning the circles would be the one generated by the curve.

Evaluate the integrals in Exercises \(41-50\) . $$ \int \frac{\operatorname{sech} \sqrt{t} \tanh \sqrt{t} d t}{\sqrt{t}} $$

Evaluate the integrals in Exercises \(41-50\) . $$ \int \tanh \frac{x}{7} d x $$

Tractor trailers and the tractrix When a tractor trailer turns into a cross street or driveway, its rear wheels follow a curve like the one shown here. (This is why the rear wheels sometimes ride up over the curb.) We can find an equation for the curve if we picture the rear wheels as a mass \(M\) at the point \((1,0)\) on the \(x\) -axis attached by a rod of unit length to a point \(P\) representing the cab at the origin. As the point \(P\) moves up the \(y\) -axis, it drags \(M\) along behind it. The curve traced by \(M-\) called a tractrix from the Latin word tractum, for "drag" - can be shown to be the graph of the function \(y=f(x)\) that solves the initial value problem $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d y}{d x}=-\frac{1}{x \sqrt{1-x^{2}}}+\frac{x}{\sqrt{1-x^{2}}}} \\ {\text { Initial condition: }} & {y=0 \quad \text { when } \quad x=1}\end{array} $$ Solve the initial value problem to find an equation for the curve. (You need an inverse hyperbolic function.) Graph cannot copy

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.