/*! 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 1 Diffirentiate. $$y=\tanh ^{2} ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

Diffirentiate. $$y=\tanh ^{2} x$$

Short Answer

Expert verified
The short answer is: \[\frac{dy}{dx} = 2\tanh x \cdot \text{sech}^2 x\].

Step by step solution

01

Differentiate \(\tanh x\)

Recall that \(\tanh x = \frac{\sinh x}{\cosh x}\). To find the derivative of \(\tanh x\), we use the quotient rule: \[\frac{d}{dx}(\tanh x) = \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{(\cosh x)^2}\] \[\frac{d}{dx}(\tanh x) = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}\] Also note that \(\cosh^2 x - \sinh^2 x = 1\), so the derivative of \(\tanh x\) is: \[\frac{d}{dx}(\tanh x) = \frac{1}{\cosh^2 x} = \text{sech}^2 x\].
02

Differentiate square function

We need to differentiate the square function \((\tanh x)^2\). Let's call \(u = \tanh x\), so we have a function \(y = u^2\). The derivative of this function with respect to \(u\) is: \[\frac{dy}{du} = 2u = 2\tanh{x}\].
03

Apply the chain rule

Now we will apply the chain rule to find the derivative \(\frac{dy}{dx}\) for \(y = (\tanh x)^2\). Recall that \(u = \tanh x\), so: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\] Substitute our results from Step 1 and Step 2: \[\frac{dy}{dx} = (2\tanh x) (\text{sech}^2 x)\]. Our final answer is: \[\frac{dy}{dx} = 2\tanh x \cdot \text{sech}^2 x\].

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.

Tanh Function
The hyperbolic tangent function, known as \( \tanh x \), is similar to the tangent function but for hyperbolic functions. It's defined as the ratio of the hyperbolic sine and hyperbolic cosine:
  • \( \tanh x = \frac{\sinh x}{\cosh x} \)
Hyperbolic functions have applications in engineering and physics, especially when dealing with wave and heat equations. Understanding \( \tanh x \) is important because it leads to other derivatives when we differentiate it.
Note that \( \tanh x \) approaches -1 and 1 as \( x \) goes to negative and positive infinity, respectively. This behavior is crucial for understanding its limits and behavior.
Quotient Rule
To differentiate a function that is the division of two other functions, like \( \tanh x = \frac{\sinh x}{\cosh x} \), you can use the quotient rule. This rule states:
  • If \( u(x) \) and \( v(x) \) are functions, then \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
For \( \tanh x \), we set \( u = \sinh x \) and \( v = \cosh x \). The derivatives \( u' = \cosh x \) and \( v' = \sinh x \) are found using basic hyperbolic derivatives.
By applying the quotient rule, we simplify to find:
  • \( \frac{d}{dx}(\tanh x) = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \text{sech}^2 x \).
Chain Rule
The chain rule is vital when dealing with composite functions, where one function is inside another. In our differentiation task, \( y = (\tanh x)^2 \) requires the chain rule because it's the tangent hyperbolic function squared.
The chain rule formula is:
  • If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
First, differentiate the outer function \( u^2 \) with respect to \( u \):
  • \( \frac{dy}{du} = 2u = 2\tanh x \).
Then multiply by the derivative of the inner function \( \tanh x \), which we've found as \( \text{sech}^2 x \):
  • \( \frac{dy}{dx} = 2\tanh x \cdot \text{sech}^2 x \).
Hyperbolic Functions
Hyperbolic functions, like \( \sinh x \), \( \cosh x \), and \( \tanh x \), are analogs of trigonometric functions but for hyperbolas rather than circles. These functions have distinctive properties and identities:
  • \( \sinh x = \frac{e^x - e^{-x}}{2} \)
  • \( \cosh x = \frac{e^x + e^{-x}}{2} \)
  • Identity: \( \cosh^2 x - \sinh^2 x = 1 \)
These identities help in simplifying derivatives and integrals involving hyperbolic functions.
Understanding these functions is essential for fields requiring hyperbolic geometry and for solving specific differential equations. They have unique asymptotic behaviors and are frequently applied in real-world scenarios, such as modeling complex systems.

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

Draw a figure that displays the graphs of both $$f(x)=\ln x \quad \text { and } \quad g(x)=\log _{2} x$$

Find all the functions \(f\) that satisfy the equation for all real \(t\). \(f^{\prime}(t)=f f(t) .\) HINT: Write \(f^{\prime}(t)-t f(t)=0\) and multiply the equation by \(e^{-t^{2} / 2}\)

Evaluate. $$\int_{0}^{1}\left(2^{x}+x^{2}\right) d x$$

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 \(g\). \(f(x)=e^{y^{y}}, \quad g(x)=\sqrt{\ln x} ; \quad x \geq 1\).

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.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

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.