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For each function, calculate "in your head" the relative rate of change. $$ D(p)=e^{-p} $$

Short Answer

Expert verified
The relative rate of change is \(-1\).

Step by step solution

01

Understand the Relative Rate of Change

The relative rate of change of a function \(D(p)\) at a point is defined as \( \frac{D'(p)}{D(p)} \). This means that we'll need to calculate the derivative \(D'(p)\), and then divide it by the original function \(D(p)\).
02

Calculate the Derivative

The given function is \(D(p) = e^{-p}\). The derivative of \(e^{-p}\) with respect to \(p\) is \(D'(p) = -e^{-p}\). This involves using the chain rule for derivatives, recognizing \(-p\) as the inner function.
03

Calculate the Relative Rate of Change

Now, calculate the relative rate of change: \[ \frac{D'(p)}{D(p)} = \frac{-e^{-p}}{e^{-p}}. \] Simplifying this gives \(-1\).
04

Interpret the Result

The relative rate of change is constant and equals \(-1\), indicating that for every unit increase in \(p\), the function \(D(p)\) decreases by a fixed percentage relative to its current value.

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Key Concepts

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

Derivative Calculation
Calculating a derivative is a fundamental aspect of understanding how functions behave. Derivatives represent the rate of change of a function with respect to a variable. For example, if we have a function \( D(p) = e^{-p} \), the derivative, denoted as \( D'(p) \), tells us how \( D(p) \) changes as \( p \) changes.

To calculate the derivative of \( D(p) = e^{-p} \), we need to apply rules of differentiation. The derivative of an exponential function like \( e^u \) is given by \( e^u \) times the derivative of \( u \) with respect to the independent variable. Therefore, for \( e^{-p} \), the derivative \( D'(p) \) becomes \( -e^{-p} \). This result uses the chain rule, since \(-p\) is an inner function that we need to differentiate.

In summary, finding the derivative is about determining the instantaneous rate at which a function changes. It is the underlying concept for understanding rates, such as velocity or growth.
Exponential Functions
Exponential functions are prevalent in many fields like science, finance, and engineering. An exponential function is characterized by its expression \( f(x) = e^x \), where \( e \) is Euler's number, an irrational constant approximately equal to 2.718. In exponential functions, the variable appears in the exponent, which leads to the rapid growth or decay.

In the original exercise, we have the function \( D(p) = e^{-p} \). This function is an example of exponential decay. The negative sign in the exponent reverses the typical growth behavior of an exponential function, causing the function's values to decrease as \( p \) increases.

Understanding exponential functions is crucial because they model a variety of real-life contexts, including population growth, radioactive decay, and interest calculations. These functions are distinguished by their consistent relative rate of change, making them valuable tools for predicting long-term trends.
Chain Rule
The chain rule is a vital technique in calculus used for differentiating compositions of functions. It connects the derivative of a composite function to the derivatives of its individual functions.

To put it simply, if you have a function \( y = f(g(x)) \), the derivative \( y' \) is found by multiplying the derivative of the outer function \( f \) evaluated at the inner function \( g(x) \), by the derivative of the inner function \( g(x) \). Mathematically, it's expressed as \( f'(g(x)) \cdot g'(x) \).

In the original problem, we used the chain rule to differentiate \( D(p) = e^{-p} \). Here, \( -p \) is the inner function, and \( e^u \) is the outer function. The derivative \( D'(p) = -e^{-p} \) is achieved by differentiating \( e^{u} \) with respect to \( u = -p \), and then multiplying by the derivative of \( u \), which is \(-1\). The chain rule is indispensable for handling more complex functions where direct differentiation isn't feasible.

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Most popular questions from this chapter

BIOMEDICAL: Ricker Recruitment The population dynamics of many fish (such as salmon) can be described by the Ricker curve \(y=\operatorname{axe}^{-b x}\) for \(x \geq 0\) where \(a>1\) and \(b>0\) are constants, \(x\) is the size of the parental stock, and \(y\) is the number of recruits (offspring). Determine the size of the parental stock that maximizes the number of recruits.

69-72. Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{x}+e^{-x} $$

Solve the following exercises on a graphing calculator by graphing an appropriate exponential function (using \(x\) for ease of entry) together with a constant function and using INTERSECT to find where they meet. You will have to choose an appropriate window. After a sale has been advertised for \(t\) days, the proportion of shoppers in a city who have seen the ad is \(1-e^{-0.08 t} .\) How long must the ad run to reach: a. \(50 \%\) of the shoppers? b. \(60 \%\) of the shoppers?

73-76. Use your graphing calculator to graph each function on the indicated interval, and give the coordinates of all relative extreme points and inflection points (rounded to two decimal places). [Hint: Use NDERIV once or twice together with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=\frac{x}{e^{x}} \text { for }-1 \leq x \leq 5 $$

77-78. Use implicit differentiation to find \(d y / d x\). $$ y^{2}-y e^{x}=12 $$

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