/*! 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 21 Differentiate. $$f(x)=-3 e^{-x... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate. $$f(x)=-3 e^{-x}$$

Short Answer

Expert verified
\( f'(x) = 3 e^{-x} \)

Step by step solution

01

Identify the Function

The given function to differentiate is \( f(x) = -3 e^{-x} \). This function is a product of a constant \(-3\) and the exponential function \( e^{-x} \).
02

Differentiate the Exponential Function

The derivative of the exponential function \( e^{-x} \) with respect to \( x \) is \( -e^{-x} \). Therefore, using this rule we have: \( \frac{d}{dx}[e^{-x}] = -e^{-x} \).
03

Apply the Constant Multiple Rule

Using the constant multiple rule, which states \( \frac{d}{dx} [cf(x)] = c \frac{d}{dx} [f(x)] \) for a constant \( c \), we can differentiate \( -3 e^{-x} \) as follows: \( \frac{d}{dx} [-3 e^{-x}] = -3 \frac{d}{dx} [e^{-x}] \).
04

Simplify the Result

Substitute the derivative of \( e^{-x} \) into the equation: \( -3 \frac{d}{dx} [e^{-x}] = -3(-e^{-x}) \). Simplify to obtain the final result: \( 3 e^{-x} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

differentiation steps
Differentiation involves specific steps to find the derivative of a function. Let's break down those steps using our example function \( f(x) = -3 e^{-x} \).

Step 1:
Identify the function to be differentiated. In this case, \( f(x) \) contains a constant multiple and an exponential function: \( -3 e^{-x} \).

Step 2:
Differentiate the exponential function while keeping the constant intact. Recall the exponential function rule: \( \frac{d}{dx}[e^{-x}] = -e^{-x} \).

Step 3:
Apply the constant multiple rule, which allows us to take the constant outside the derivative operation:
  • Use \( -3 \frac{d}{dx} [e^{-x}] \)
Step 4:
Substitute the derivative of the exponential function from Step 2: \( -3 [-e^{-x}] \). Simplify the result to get the final derivative:

Step 4:
Final simplification gives you \( 3 e^{-x} \). These structured steps make differentiation systematic and easier to follow.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We have now studied models for linear, quadratic, exponential, and logistic growth. In the real world, understanding which is the most appropriate type of model for a given situation is an important skill. Identify the most appropriate type of model and explain why you chose that model. List any restrictions you would place on the domain of the function. The growth in value of a U.S. savings bond

A quantity \(Q_{1}\) grows exponentially with a doubling time of 1 yr. A quantity \(Q_{2}\) grows exponentially with a doubling time of 2 yr. If the initial amounts of \(Q_{1}\) and \(Q_{2}\) are the same, how long will it take for \(Q_{1}\) to be twice the size of \(Q_{2} ?\)

Suppose that \(\$ 100\) is invested at \(7 \%,\) compounded continuously, for 1 yr. We know from Example 4 that the ending balance will be \(\$ 107.25 .\) This would also be the ending balance if \(\$ 100\) were invested at 7.25 \(\%,\) compounded once a year (simple interest). The rate of \(7.25 \%\) is called the effective annual yield. In general, if \(P_{0}\) is invested at interest rate \(k,\) compounded continuously, then the effective annual yield is that number i satisfying \(P_{0}(1+i)=P_{0} e^{k} .\) Then, \(1+i=e^{h},\) or Effective annual yield \(=i=e^{k}-1\) An amount is invested at \(8 \%\) per year compounded continuously. What is the effective annual yield?

Suppose that \(\$ 100\) is invested at \(7 \%,\) compounded continuously, for 1 yr. We know from Example 4 that the ending balance will be \(\$ 107.25 .\) This would also be the ending balance if \(\$ 100\) were invested at 7.25 \(\%,\) compounded once a year (simple interest). The rate of \(7.25 \%\) is called the effective annual yield. In general, if \(P_{0}\) is invested at interest rate \(k,\) compounded continuously, then the effective annual yield is that number i satisfying \(P_{0}(1+i)=P_{0} e^{k} .\) Then, \(1+i=e^{h},\) or Effective annual yield \(=i=e^{k}-1\) An amount is invested at \(7.3 \%\) per year compounded continuously. What is the effective annual yield?

Use input-output tables to find each limit. $$\lim _{x \rightarrow \infty} \ln x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.