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Differentiate. $$ y=1-e^{-k x} $$

Short Answer

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The derivative is \( y' = ke^{-kx} \).

Step by step solution

01

Apply Basic Differentiation Rules

To differentiate the function, we will apply standard differentiation rules. The function is given as \( y = 1 - e^{-kx} \). The derivative of a constant, like 1, is zero. Hence, we focus on differentiating \( -e^{-kx} \).
02

Use the Chain Rule

The function \( e^{-kx} \) requires the chain rule because it is an exponential function with an inner function \( -kx \). The derivative of \( e^u \) with respect to \( u \) is \( e^u \), and when differentiating \( -kx \) with respect to \( x \), we get \( -k \).
03

Differentiate \(-e^{-kx}\)

Following the chain rule, the derivative of \(-e^{-kx}\) is \((-1)\cdot (e^{-kx}) \cdot (-k)\). Simplifying this, we get \( ke^{-kx} \).
04

Compile the Results

Combine the results of differentiating each part. The derivative of \( 1 \) is 0 and \( -e^{-kx} \) is \( ke^{-kx} \). Thus, the derivative of the entire function \( y \) is \( y' = 0 + ke^{-kx} = ke^{-kx} \).

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Key Concepts

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

Understanding the Chain Rule
Differentiation is a fundamental concept in calculus that allows us to determine the rate at which functions change. When dealing with complex functions, we often use a technique known as the Chain Rule. The Chain Rule is essential when we have composite functions, meaning a function within another function. In simpler terms, it helps us differentiate functions that are nested inside each other.

Here's how the chain rule works:
  • Identify the outer and inner functions. For example, in the function \( e^{-kx} \), the outer function is \( e^u \) and the inner function is \( -kx \).
  • Differentiating the outer function takes place first, treating the inner function as a mere placeholder. The derivative of \( e^u \) with respect to \( u \) is \( e^u \).
  • Next, differentiate the inner function. The derivative of \( -kx \) with respect to \( x \) is \( -k \).
  • Finally, multiply the derivatives of the inner and outer functions. In our case, this results in \( e^{-kx} \cdot (-k) \).
By systematically applying these steps, the chain rule allows us to confidently tackle the differentiation of nested functions.
Exploring Exponential Functions
Exponential functions are ubiquitous in mathematics and applicable in various real-world phenomena such as population growth and radioactive decay. An exponential function typically involves a constant base raised to a variable exponent. For instance, \( e^{-kx} \) is an exponential function.

Exponential functions with a base of \( e \) (approximately 2.718) are particularly significant in calculus due to a special property: they are their own derivatives. This means that when you differentiate \( e^x \), you get \( e^x \) again. This property simplifies the process of differentiation and highlights why base \( e \) is extensively used.
  • When differentiating \( e^{ax} \), where \( a \) is a constant, the derivative is \( ae^{ax} \). This pattern makes it easy to predict how such functions change over time or space.
  • In our problem, the differentiation of \( e^{-kx} \) also involves applying existing differentiation rules to the exponent \( -kx \).
Being familiar with exponential functions and their properties significantly eases understanding and solving calculus problems involving growth, decay, and oscillations.
Applying Basic Differentiation Rules
When learning differentiation, basic differentiation rules form the foundation upon which more complex rules such as the chain rule are built. These simple rules enable us to differentiate elementary functions quickly and accurately.

Some of the fundamental rules include:
  • The derivative of any constant is zero.
  • The derivative of \( x^n \), where \( n \) is a constant, is \( nx^{n-1} \).
  • The derivative of \( c\cdot x \), where \( c \) is a constant, is simply \( c \).
Applying these rules to our original function \( y = 1 - e^{-kx} \):
  • The derivative of the constant \( 1 \) is zero, so it does not affect the differentiation of the entire function.
Understanding these foundations allows us to break down the problem and tackle it part by part. It simplifies the approach to increasingly challenging calculus concepts.

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Most popular questions from this chapter

The Rule of 70. The relationship between doubling time \(T\) and growth rate \(k\) is the basis of the Rule of \(70 .\) Since $$ T=\frac{\ln 2}{k}=\frac{0.693147}{k}=\frac{69.3147}{100 k} \approx \frac{70}{100 k} $$ we can estimate the length of time needed for a quantity to double by dividing the growth rate \(k\) (expressed as a percentage) into \(70 .\) Estimate the time needed for an amount of money to double, if the interest rate is \(7 \%,\) compounded.

More Americans are buying organic fruit and vegetables and products made with organic ingredients. The amount \(A(t)\), in billions of dollars, spent on organic food and beverages \(t\) years after 1995 can be approximated by $$ A(t)=2.43 e^{0.18 t} $$ a) Estimate the amount that Americans spent on organic food and beverages in \(2009 .\) b) Estimate the rate at which spending on organic food and beverages was growing in \(2006 .\)

Bornstein and Bornstein found in a study that the average walking speed \(v,\) in feet per second, of a person living in a city of population \(p,\) in thousands, is $$v(p)=0.37 \ln p+0.05 $$. a) The population of Seattle is \(635,000(p=635)\). What is the average walking speed of a person living in Seattle? b) The population of New York is \(8,340,000 .\) What is the average walking speed of a person living in New York? c) Find \(v^{\prime}(p)\). d) Interpret \(v^{\prime}(p)\) found in part (c).

Graph \(f, f^{\prime},\) and \(f^{\prime \prime}\) $$ f(x)=e^{-x} $$

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?

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