/*! 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 12 Verify the formula. $$\frac{d}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 12

Verify the formula. $$\frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x$$

Short Answer

Expert verified
We find the derivative of \(\operatorname{sech}(x)\) by using the chain rule on the function \(\frac{1}{\cosh(x)}\). First, we calculate the derivatives of \(f(x) = x^{-1}\) and \(g(x) = \cosh(x)\) as \(f'(x) = \frac{-1}{x^2}\) and \(g'(x) = \sinh(x)\). Applying the chain rule, we get the derivative as: \(-\frac{\sinh(x)}{\cosh(x)}\cdot\frac{1}{\cosh(x)} = -\tanh(x) \operatorname{sech}(x)\). Thus, the given formula has been verified.

Step by step solution

01

Recall the hyperbolic identities

We need to remember the identity for the hyperbolic secant and hyperbolic tangent functions: \[\operatorname{sech}(x) = \frac{1}{\cosh(x)}\] \[\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\]
02

Calculate the derivative of the hyperbolic secant function

We know that \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\), so we need to find the derivative of this function with respect to \(x\): \[ \frac{d}{d x}\left(\frac{1}{\cosh(x)}\right)\]
03

Use the chain rule

To find the derivative of \(\frac{1}{\cosh(x)}\) with respect to \(x\), we will use the chain rule. The chain rule states that \(\frac{d}{d x}(f(g(x))) = f'(g(x))g'(x)\), where \(f\) and \(g\) are differentiable functions. Let \(f(x) = x^{-1}\), and \(g(x) = \cosh(x)\). Then, the function we're trying to find the derivative of can be written as \(f(g(x))\), and we can apply the chain rule. First, we need to find the derivatives of \(f\) and \(g\): \[f'(x) = \frac{-1}{x^2}\] \[g'(x) = \sinh(x)\] Now we can apply the chain rule: \[\frac{d}{d x}\left(\frac{1}{\cosh(x)}\right) = f'(g(x))g'(x) = \frac{-1}{(\cosh(x))^2}\sinh(x)\]
04

Express the derivative using hyperbolic identities

Using the hyperbolic identities from Step 1, we can write the derivative we found in Step 3 using the original hyperbolic functions: \[\frac{-1}{(\cosh(x))^2}\sinh(x) = -\frac{\sinh(x)}{\cosh(x)}\cdot\frac{1}{\cosh(x)} = -\tanh(x) \operatorname{sech}(x)\]
05

Compare with the given formula

We needed to verify that \(\frac{d}{d x}(\operatorname{sech}(x)) = -\operatorname{sech}(x) \tanh(x)\). From the calculation in Step 4, we find that the derivative of the hyperbolic secant function is exactly equal to the given formula, so the formula is verified: \[\frac{d}{d x}(\operatorname{sech}(x)) = -\operatorname{sech}(x) \tanh(x)\]

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.

Hyperbolic Functions
Hyperbolic functions can be likened to trigonometrical functions but for hyperbolic shapes. They include functions like hyperbolic sine (\( \sinh(x) \)) and hyperbolic cosine (\( \cosh(x) \)), which are central to calculus involving exponential growth and decay.
  • The hyperbolic secant (\( \operatorname{sech}(x) \)) is defined as \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \), symbolizing the cosine's inverse concerning the hyperbolic context.
  • The hyperbolic tangent (\( \tanh(x) \)) is expressed as \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \), modeling the sine's ratio over cosine in a hyperbolic version.
Unlike regular trigonometric functions, hyperbolic functions are highly relevant in fields such as physics, particularly in the areas dealing with special relativity and Einstein's theory of gravitation.
These functions are smoother and continuous, extending real calculations into more advanced realms.
Chain Rule
The chain rule is a fundamental aspect of calculus that allows us to differentiate composite functions. When you have a function composed of another function, use the chain rule to break it down into manageable parts.
  • The chain rule is formulated as: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This means you take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function.
  • In our context, we had \( f(x) = x^{-1} \) and \( g(x) = \cosh(x) \). Knowing their derivatives, \( f'(x) = \frac{-1}{x^2} \) and \( g'(x) = \sinh(x) \), helps us apply the chain rule effectively.
Applying the chain rule results in simplifying complex derivatives, allowing easier verification and manipulation of hyperbolic functions in calculus.
Proof Verification
Proof verification in calculus involves checking if mathematical identities or derivatives are correctly derived and if they are consistent with foundational principles. In our case, we had to verify:
\[\frac{d}{d x}(\operatorname{sech}(x))=-\operatorname{sech}(x) \tanh(x)\]
Many steps are usually involved:
  • First, recall the known identities, such as those of hyperbolic functions you've previously learned.
  • Next, compute the derivative explicitly and directly, as was done step-by-step above.
  • Finally, compare your derived expression with the one you want to confirm.
If the expression matches, as it did here, it confirms the theory and approach were correct, thereby solidifying your understanding of both hyperbolic functions and their derivatives.
Developing this skill empowers you to strengthen any mathematical matter, providing clarity and precision.

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

Find the crilical points and the extreme values. Taxe \(k\) as a positive integer. (a) \(f(x)=x^{k} \ln x, \quad x>0\). (b) \(f(x)=x^{k} e^{-x}, \quad x\) real.

The compound interest formula $$Q=P e^{r^{t}}$$ can be written $$P=Q e^{-r t}$$ In this formulation we have \(P\) as the investment required today to obtain \(Q\) in \(t\) years. In this sense \(P\) dollars is present value of \(Q\) dollars to be received \(t\) years from now. Find the present value of 20,000 dollar to be received 4 years from now. Assume continuous compounding at \(4 \%\).

Use a graphing utility to draw a figure that displays the graphs of \(f\) and \(g\). The figure should suggest that \(f\) and \(g\) are inverses. Show that this is true by verifying that \(f(g(x))=x\) for each \(x\) in the domain of \(g\). \(f(x)=e^{2 x}, \quad g(x)=\ln \sqrt{x} ; \quad x>0\).

Estimate the limit numerically by evaluating the function at the indicated values of \(x\). Then use a graphing utility to zoom in on the graph and justify your estimate. $$\begin{aligned} &\lim _{x \rightarrow 1} \frac{\ln x}{x-1} ; \quad x=1 \pm 0.5 ,1 \pm 0.1,1 \pm 0.01,1 \pm 0.001,1 \pm 0.0001\end{aligned}$$

Let \(P\) be a polynomial of degree \(n\) (a) Can \(P\) have an inverse if \(n\) is even? Support your answer. (b) Can \(P\) have an inverse if \(n\) is odd? If so, give an example. Then give an example of a polynomial of odd degree that does not have an inverse.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Discrete 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.