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Chapter 3: Problem 18

Differentiate. $$ F(x)=e^{-4 x} $$

Short Answer

Expert verified
The derivative is \(-4e^{-4x}\).

Step by step solution

01

Understand the Exponential Function

The function provided is an exponential function where the base is the natural exponent, \( e \), and the exponent is \( -4x \). Thus, \( F(x) = e^{-4x} \) represents a function that decreases exponentially as \( x \) increases.
02

Apply the Chain Rule

To differentiate a function of the form \( e^{u} \), where \( u \) is a differentiable function of \( x \), we use the chain rule. The derivative is \( e^{u} \cdot \frac{du}{dx} \). In this case, \( u = -4x \).
03

Differentiate the Inner Function

Find the derivative of the inner function \( u = -4x \). This is a simple linear function, and the derivative with respect to \( x \) is \( \frac{d(-4x)}{dx} = -4 \).
04

Calculate the Derivative

Apply the chain rule using the results from the previous steps. The derivative of \( F(x) = e^{-4x} \) is given by:\[ F'(x) = e^{-4x} \cdot (-4) \]Thus, the derivative simplifies to:\[ F'(x) = -4e^{-4x} \]

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Key Concepts

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

Exponential Functions
Exponential functions are a key concept in calculus and are incredibly useful in various fields such as science, engineering, and economics. An exponential function is a function of the form \(f(x) = a \cdot e^{bx}\), where \(e\) is the base of natural logarithms, approximately equal to 2.71828. These functions exhibit unique growth and decay properties:

  • Growth/Decay Behavior: If the exponent \(b\) is positive, the function represents exponential growth. If \(b\) is negative, it signifies exponential decay.
  • Characteristics: Exponential functions have a constant ratio. This means that as you move along the x-axis, the function's value changes by a constant multiplier rather than a constant increment.

In the function \(F(x) = e^{-4x}\), the negative exponent \(-4x\) indicates an exponential decay, meaning as \(x\) increases, \(F(x)\) decreases.
Chain Rule
The chain rule is a powerful tool in calculus used to differentiate composite functions. When dealing with functions that have other functions nested inside, like \(F(x) = e^{-4x}\), the chain rule becomes essential. It helps in breaking down the differentiation into manageable steps:

  • Application: To use the chain rule, identify the outer function and the inner function. Differentiate both separately, then multiply them together.
  • Formula: If you have a function \(f(g(x))\), the derivative is calculated as \(f'(g(x)) \cdot g'(x)\).

In our example, \(e^u\) is the outer function and \(u = -4x\) is the inner function.
By differentiating the outer function \(e^u\) and the inner function \(-4x\), we effectively apply the chain rule, simplifying the differentiation process.
Derivative Calculation
Calculating the derivative is the final step after understanding the function type and applying the chain rule. Derivatives measure the rate of change of a function concerning one of its variables and are foundational in calculus for understanding behavior of functions:

  • Inner Function Derivative: Differentiate the inner function \(u = -4x\). This is linear, so its derivative is simply \(-4\).
  • Outer Function Derivative: The derivative of \(e^{u}\) with respect to \(u\) is \(e^{u}\) itself.

Now, multiply these results to find the derivative of the original function. For \(F(x) = e^{-4x}\), the derivative becomes:
\[F'(x) = e^{-4x} \cdot (-4) = -4e^{-4x}\]
This shows how the chain rule can simplify what initially seems complex, resulting in a manageable expression that describes how \(F(x)\) changes as \(x\) increases.

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Most popular questions from this chapter

A beam of light enters a medium such as water or smoky air with initial intensity \(I_{0}\). Its intensity is decreased depending on the thickness (or concentration) of the medium. The intensity I at a depth (or concentration) of \(x\) units is given by $$I=I_{0} e^{-\mu x}$$ The constant \(\mu\left({ }^{u} m u "\right),\) called the coefficient of absorption, varies with the medium. Use this law for Exercises 62 and \(63 .\) Concentrations of particulates in the air due to pollution reduce sunlight. In a smoggy area, \(\mu=0.01\) and \(x\) is the concentration of particulates measured in micrograms per cubic meter \(\left(\mathrm{mcg} / \mathrm{m}^{3}\right)\) What change is more significant-dropping pollution levels from \(100 \mathrm{mcg} / \mathrm{m}^{3}\) to \(90 \mathrm{mcg} / \mathrm{m}^{3}\) or dropping them from \(60 \mathrm{mcg} / \mathrm{m}^{3}\) to \(50 \mathrm{mcg} / \mathrm{m}^{3}\) ? Why?

As part of a study, students in a psychology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After \(t\) months, the average score \(S(t),\) as a percentage, was found to be given by $$ S(t)=78-15 \ln (t+1), \quad t \geq 0 $$. a) What was the average score when they initially took the test, \(t=0 ?\) b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years ( 24 months)? e) Find \(S^{\prime}(t)\). f) Find the maximum and minimum values, if they exist. g) Find \(\lim _{t \rightarrow \infty} S(t)\) and discuss its meaning.

Differentiate. $$ f(x)=\frac{x e^{-x}}{1+x^{2}} $$

At a price of \(x\) dollars, the demand, in thousands of units, for a certain turntable is given by the demand function $$ q=240 e^{-0.003 x} $$ a) How many turntables will be bought at a price of \(\$ 250 ?\) Round to the nearest thousand. b) Graph the demand function for \(0 \leq x \leq 400\). c) Find the marginal demand, \(q^{\prime}(x)\). d) Interpret the meaning of the derivative.

An interest rate decreases from \(8 \%\) to \(7.2 \% .\) Explain why this increases the present value of an amount due 10 yr later.
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