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Chapter 5: Problem 17

The hyperbolic functions are defined as \(\sinh x=\frac{1}{2}\left(e^{x}-e^{-x}\right)\) and \(\cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right)\). a. Prove \(\frac{d(\sinh x)}{d x}=\cosh x\). b. Prove \(\frac{d(\cosh x)}{d x}=\sinh x\). c. Prove \(\frac{d(\tanh x)}{d x}=\frac{1}{(\cosh x)^{2}}\) if \(\tanh x=\frac{\sinh x}{\cosh x}\).

Short Answer

Expert verified
a. \( \frac{d(\sinh x)}{dx} = \cosh x \). b. \( \frac{d(\cosh x)}{dx} = \sinh x \). c. \( \frac{d(\tanh x)}{dx} = \frac{1}{(\cosh x)^2} \)."

Step by step solution

01

Differentiate sinh(x)

The function is defined as \( \sinh x = \frac{1}{2} \left(e^x - e^{-x}\right) \). Differentiate it with respect to \( x \):\[ \frac{d(\sinh x)}{dx} = \frac{1}{2} \left( \frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x}) \right) \]Applying the derivative rules, we have:\[ \frac{d}{dx}(e^x) = e^x \quad \text{and} \quad \frac{d}{dx}(e^{-x}) = -e^{-x} \]Thus,\[ \frac{d(\sinh x)}{dx} = \frac{1}{2} (e^x + e^{-x}) = \cosh x \].
02

Differentiate cosh(x)

The function is defined as \( \cosh x = \frac{1}{2} \left(e^x + e^{-x}\right) \). Differentiate it with respect to \( x \):\[ \frac{d(\cosh x)}{dx} = \frac{1}{2} \left( \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{-x}) \right) \]Using the previous derivatives:\[ \frac{d}{dx}(e^x) = e^x \quad \text{and} \quad \frac{d}{dx}(e^{-x}) = -e^{-x} \]Thus,\[ \frac{d(\cosh x)}{dx} = \frac{1}{2} (e^x - e^{-x}) = \sinh x \].
03

Differentiate tanh(x) using the quotient rule

The function is defined as \( \tanh x = \frac{\sinh x}{\cosh x} \). Use the quotient rule for derivatives, which states \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = \sinh x \) and \( v = \cosh x \).From previous steps:\[ u' = \cosh x \quad \text{and} \quad v' = \sinh x \]Apply the quotient rule:\[ \frac{d(\tanh x)}{dx} = \frac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{(\cosh x)^2} \]Utilize the identity \( \cosh^2 x - \sinh^2 x = 1 \):\[ \frac{d(\tanh x)}{dx} = \frac{1}{(\cosh x)^2} \].

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Key Concepts

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

Differential Calculus
Differential calculus is a fundamental branch of mathematics that deals with the concept of change. Specifically, it focuses on how functions change when their inputs vary. A key element in differential calculus is the derivative, which is essentially the rate of change or the slope of a function at any given point.
In more formal terms, if we have a function \( y = f(x) \), then its derivative \( \frac{dy}{dx} \) represents how much \( y \) changes with a small change in \( x \). This concept is crucial in understanding and analyzing real-world phenomena where variables are not static but change in relation to each other, like velocity, acceleration, and dynamic systems.
In the context of our problem, we're dealing with derivatives of hyperbolic functions鈥攕pecial functions that have applications in many fields such as engineering, physics, and mathematics. Understanding how to derive these functions is a significant part of differential calculus.
Derivatives of Hyperbolic Functions
Hyperbolic functions are analogous to the trigonometric functions, but they are based on hyperbolas instead of circles. The derivatives of these functions have unique properties that make them fascinating to explore.
The hyperbolic sine function, \( \sinh x \), is defined as \( \frac{1}{2} \left(e^x - e^{-x}\right) \). Its derivative, as derived in the exercise, is the hyperbolic cosine function, \( \cosh x \), defined as \( \frac{1}{2} \left(e^x + e^{-x}\right) \). Similarly, the derivative of the hyperbolic cosine function is the hyperbolic sine function itself, which means these functions are deeply interconnected.
These properties arise from the exponential nature of \( e^x \), which makes them consistently recur across derivatives. Being familiar with hyperbolic functions and their derivatives is essential because they frequently appear in mathematical modeling and analysis of physical processes.
Quotient Rule
The quotient rule is a bit like a strategic tool in differential calculus when dealing with derivatives of functions in the form of a quotient鈥攖hat is, one function divided by another.
The quotient rule formula is \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u \) and \( v \) are differentiable functions of \( x \). This rule plays a pivotal role in finding the derivative of \( \tanh x \), where \( \tanh x = \frac{\sinh x}{\cosh x} \).
In this exercise, we showed that the derivative of \( \tanh x \) is \( \frac{1}{(\cosh x)^2} \) using the quotient rule. By applying it, we saw the interaction between the derivatives of \( \sinh x \) and \( \cosh x \) produce a simple yet powerful expression. Mastering the quotient rule and understanding how to apply it to complex functions is an invaluable skill in calculus, allowing one to tackle a wide array of differential problems efficiently.

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Most popular questions from this chapter

The squirrel population in a small self-contained forest was studied by a biologist. The biologist found that the squirrel population, \(P\), measured in hundreds, is a function of time, \(t,\) where \(t\) is measured in weeks. The function is \(P(t)=\frac{20}{1+3 e^{-0.02 t}}\) a. Determine the population at the start of the study, when \(t=0\) b. The largest population the forest can sustain is represented mathematically by the limit as \(t \rightarrow \infty .\) Determine this limit. c. Determine the point of inflection. d. Graph the function. e. Explain the meaning of the point of inflection in terms of squirrel population growth.

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