/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Let \(y(x)=e^{x}\). Compute \(y^... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(y(x)=e^{x}\). Compute \(y^{\prime}(x), y^{\prime \prime}(x)=\left(y^{\prime}\right)^{\prime},\) and \(y^{\prime \prime \prime}(x)\).

Short Answer

Expert verified
The derivatives are: \(y'(x) = e^x\), \(y''(x) = e^x\), \(y'''(x) = e^x\).

Step by step solution

01

Function Identification

Identify the given function, which is \(y(x) = e^x\). This function will be differentiated multiple times as per the problem requirement.
02

First Derivative - \(y'(x)\)

Find the first derivative \(y'(x)\) using the fact that the derivative of \(e^x\) with respect to \(x\) is \(e^x\).\[ y'(x) = \frac{d}{dx}(e^x) = e^x \]
03

Second Derivative - \(y''(x)\)

Differentiate \(y'(x) = e^x\) again to find the second derivative \(y''(x)\). Since the derivative of \(e^x\) is still \(e^x\), we have:\[ y''(x) = \frac{d}{dx}(e^x) = e^x \]
04

Third Derivative - \(y'''(x)\)

Differentiate \(y''(x) = e^x\) again to find the third derivative \(y'''(x)\). Once more, the derivative of \(e^x\) equals \(e^x\), so:\[ y'''(x) = \frac{d}{dx}(e^x) = e^x \]

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Key Concepts

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

Differentiation
Differentiation is a key concept in calculus that involves finding the rate at which a function changes at any given point. This process is fundamentally about determining the derivative, which is the slope of the tangent line to the curve of the function at a particular point. In the exercise, we are differentiating the exponential function \(y(x) = e^x\), which is a fundamental process for understanding how functions behave and change.
The differentiation process applies to not only finding the first derivative but can also extend to higher-order derivatives like the second or third. In practical terms, the first derivative can tell you the function's rate of change, while higher-order derivatives can provide insights into the function's curvature and concavity, as well as motion dynamics.
  • The first derivative, often denoted by \(y'(x)\), gives the slope or rate of change.
  • Second and third derivatives, \(y''(x)\) and \(y'''(x)\), give information about how this rate itself is changing, such as identifying acceleration in physical contexts.
Understanding differentiation allows for deeper insights into mathematical modeling and problem-solving in both pure and applied mathematics.
Exponential Functions
Exponential functions are a class of functions characterized by a constant base raised to a variable exponent, and \(y(x) = e^x\) is one of the most common examples. These functions are essential in mathematics because they grow at rates proportional to their current value, which makes them suitable for modeling real-world phenomena like population growth, radioactive decay, and compound interest.
The base of the exponential function \(e^x\) is Euler's number \(e\), approximately equal to 2.718. What makes \(e^x\) unique is its property that its derivative is the same as the original function itself. This property proves incredibly useful in calculus because it simplifies the analysis of rates of change.
  • Exponential growth or decay is represented succinctly through these functions.
  • This property of self-derivation is critical in solving differential equations, which appear extensively in fields like physics and engineering.
The intuitive understanding of exponential functions allows students and professionals to accurately model dynamic systems and predict future behaviors.
Derivative Rules
Derivative rules are guidelines or formulas used to make the process of finding derivatives more systematic and less prone to error. For the exponential function \(y = e^x\), we use the rule that states the derivative of \(e^x\) with respect to \(x\) is simply \(e^x\) itself.
These rules are vital because they provide a structured approach to differentiation, allowing us to quickly find rates of change without computing limits from first principles each time.
  • The power rule, chain rule, and product rule are some of the most commonly used derivative rules.
  • While ' extit{e} extsuperscript{x}' maintains its form upon differentiation, other functions may require using these rules to resolve more complex differentiations.
  • Understanding these rules paves the way for solving more intricate calculus problems efficiently and accurately.
Each rule serves as a building block, allowing budding mathematicians to explore more advanced concepts with confidence and proficiency.

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Most popular questions from this chapter

Draw the graphs of $$y(t)=e^{t} \quad \text { and } \quad p(t)=1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}+\frac{t^{4}}{24}$$ Set the domain and range to \(-1 \leq t \leq 2\). and \(0 \leq y \leq 8\).

Suppose penicillin concentration in the serum of a patient \(t\) minutes after a bolus injection of \(2 \mathrm{~g}\) is given by $$P(t)=200 \times 0.96^{t} \quad \mu \mathrm{g} / \mathrm{ml}$$ a. Approximate \(P(t)\) and \(P^{\prime}(t)\) for \(t=0,5,10,15,\) and 20 minutes. b. Plot a graph of \(P^{\prime}(t)\) vs \(P(t)\) using the five pairs of values you just computed.

The function \(b^{t}\) for \(b=1\) is a special exponential function. Confirm that the derivative equation \(\left[b^{t}\right]^{\prime}=b^{t} \ln b\) is valid for \(b=1\). Draw some graphs of \(b^{t}\) for \(b=1\) and its derivative.

Consider the kinetics of penicillin that is taken as a pill in the stomach. The diagram in Figure Ex. \(5.4 .9(\) a \()\) may help visualize the kinetics. We will find in Chapter 17 that a model of plasma concentration of antibiotic \(t\) hours after ingestion of an antibiotic pill yields an equation similar to $$C(t)=5 e^{-2 t}-5 e^{-3 t} \quad \mu \mathrm{g} / \mathrm{ml}$$ A graph of \(C\) is shown in Figure Ex. \(5.4 .9 .\) At what time will the concentration reach a maximum level, and what is the maximum concentration achieved? As we saw in Section 3.5 .2 and may be apparent from the graph in Figure Ex. 5.4 .9 , the highest concentration is associated with the point of the graph of \(C\) at which \(C^{\prime}=0 ;\) the tangent at the high point is horizontal. The question, then, is at what time \(t\) is \(C^{\prime}(t)=0\) and what is \(C(t)\) at that time?

Plasma penicillin concentration is $$P(t)=5 e^{-0.3 t}-5 e^{-0.4 t}$$ \(t\) hours after ingestion of a penicillin pill into the stomach. A small amount of the drug diffuses into tissue and the tissue concentration, \(C(t)\), is $$ C(t)=-e^{-0.3 t}+0.5 e^{-0.4 t}+0.5 e^{-0.2 t} \quad \mu \mathrm{g} / \mathrm{ml} $$ a. Use your technology (calculator or computer) to find the time at which the concentration of the drug in tissue is maximum and the value of \(C\) at that time. b. Compute \(C^{\prime}(t)\) and solve for \(t\) in \(C^{\prime}(t)=0\). This is really bad, for you must solve for \(t\) in $$ 0.3 e^{-0.3 t}-0.2 e^{-0.4 t}-0.1 e^{-0.2 t}=0 $$ Try this: $$ \text { Let } \quad Z=e^{-0.1 t} \quad \text { then solve } \quad 0.3 Z^{3}-0.2 Z^{4}-0.1 Z^{2}=0 . $$ c. Solve for the possible values of \(Z\). Remember that \(Z=e^{-0.1 t}\) and solve for \(t\) if possible using the possible values of \(Z\). d. Which value of \(t\) solves our problem?

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