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Chapter 12: Problem 19

Find the derivative of the given function.\(f(x)=e^{x} \cosh x\)

Short Answer

Expert verified
The derivative of \( f(x) = e^{x} \cosh x \) is \( f'(x) = e^{x}(\cosh x + \sinh x) \)

Step by step solution

01

Identify the Function Components

The function given is a product of two functions: 1. The exponential function: \(e^{x}\)2. The hyperbolic cosine function: \( \cosh x \)
02

Apply the Product Rule

To find the derivative of the product of two functions, use the product rule: \[ (uv)' = u'v + uv' \]Here, \( u = e^{x} \) and \( v = \cosh x \)
03

Differentiate the Exponential Function

Differentiate \[ u = e^{x} \] with respect to \ x \The derivative is: \[ u' = e^{x} \]
04

Differentiate the Hyperbolic Cosine Function

Differentiate \[ v = \cosh x \] with respect to \ x \The derivative is: \[ v' = \sinh x \]
05

Apply the Product Rule Formula

Now substitute \ u, v, u' \ and \ v' \ into the product rule formula: \[ (e^{x} \cosh x)' = e^{x} \sinh x + e^{x} \cosh x \]
06

Combine the Terms

Combine the terms to get the final answer: \[ f'(x) = e^{x}(\cosh x + \sinh x) \]

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Key Concepts

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

Product Rule
When dealing with the derivative of a product of two functions, we use the product rule. The product rule states that if you have two functions, say \(u(x)\) and \(v(x)\), their derivative is given by \[ (uv)' = u'v + uv' \]. This means you need to:
  • Differentiate the first function while keeping the second function as it is
  • Then add the product of the first function as it is and the derivative of the second function
You must understand how crucial it is to apply this rule correctly for finding the derivatives of products of functions.
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for the hyperbola. Examples include the hyperbolic cosine \(\text{cosh}(x)\) and the hyperbolic sine \(\text{sinh}(x)\). These functions have special derivatives:
  • The derivative of \(\text{cosh}(x)\) is \(\text{sinh}(x)\)
  • The derivative of \(\text{sinh}(x)\) is \(\text{cosh}(x)\)
Understanding these derivatives is essential when dealing with problems involving hyperbolic functions, as they often appear in calculus and differential equations.
This symmetry between the functions is one of the interesting properties that make hyperbolic functions unique and useful in various applications.
Exponential Functions
Exponential functions, like \(e^x\), are functions where the variable appears as an exponent. One of the unique properties of the exponential function \(e^x\) is that its derivative is itself: \(\frac{d}{dx} e^x = e^x\).

This means that when you differentiate \(e^x\), the result is simply \(e^x\).
This property makes calculations involving exponential functions much simpler and is why they appear frequently in mathematical problems.

Exponential functions are widely used in various fields including physics, engineering, and economics, due to their properties of growth and decay.

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Most popular questions from this chapter

Find the derivative of the given function.\(f(x)=x^{\sinh x}\)

Express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.\(\int \frac{x d x}{\sqrt{x^{4}-1}}\)

Express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.\(\int \frac{d x}{\sqrt{x^{2}-4 x+1}}\)

Find the derivative of the given function.\(f(x)=\sinh ^{-1}(\tan x)\)

Find the derivative of the given function.\(f(x)=\ln (\operatorname{coth} 3 x-\operatorname{csch} 3 x)\)
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