/*! 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 26 Differentiate the functions in P... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate the functions in Problems. Assume that \(A\) \(B,\) and \(C\) are constants. $$y=t^{2}+5 \ln t$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dt} = 2t + \frac{5}{t} \).

Step by step solution

01

Identify the Terms to Differentiate

The given function is \( y = t^2 + 5 \ln t \). We need to differentiate each term separately. The first term is \( t^2 \), a power function, and the second term is \( 5 \ln t \), which is a logarithmic function multiplied by a constant.
02

Differentiate the First Term

Differentiate \( t^2 \) using the power rule, which states that \( \frac{d}{dt}(t^n) = n \cdot t^{n-1} \). For \( t^2 \), \( n = 2 \), so the derivative is \( 2 \cdot t^{2-1} = 2t \).
03

Differentiate the Second Term

For the term \( 5 \ln t \), apply the constant multiple and logarithmic differentiation rules. The derivative of \( \ln t \) is \( \frac{1}{t} \), thus the derivative of \( 5 \ln t \) is \( 5 \cdot \frac{1}{t} = \frac{5}{t} \).
04

Combine the Derivatives

Now, combine the derivatives of the two terms to find \( \frac{dy}{dt} \). The derivative of the function \( y = t^2 + 5 \ln t \) is \( \frac{dy}{dt} = 2t + \frac{5}{t} \).

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.

Power Rule
The power rule is a fundamental tool in calculus, used to differentiate functions that are powers of variables. Basically, it’s a shortcut that helps us find the derivative of functions like \( t^2 \). The power rule states that if you have a function like \( f(t) = t^n \), its derivative \( f'(t) \) is \( n \cdot t^{n-1} \).
For example, to differentiate \( t^2 \):
- Identify \( n \), which is 2 in this case.- Apply the power rule by multiplying \( n = 2 \) with \( t^{n-1} = t^{1} \).
This means the derivative is \( 2t \).
Using the power rule is often the quickest way to find the derivative of polynomial functions. It makes handling terms like quadratic and cubic much easier. Understanding how to pinpoint \( n \) from the function and applying the reduction in exponent is key for mastering this rule. As you get comfortable, you'll find that it saves a lot of time compared to using the basic definition of a derivative.
Logarithmic Differentiation
Logarithmic differentiation is a special technique used to differentiate functions with logarithmic expressions. It's especially handy for functions where a variable is inside a logarithm. For a basic function \( f(t) = \ln t \), the derivative is \( \frac{1}{t} \). When a constant is multiplied by the logarithm, like \( 5 \ln t \), the constant rule applies.
  • The constant rule states that when differentiating a constant times a function, you can "carry over" the constant.
  • In our case, differentiate \( \ln t \) which gives \( \frac{1}{t} \).
  • Then multiply it by the constant \( 5 \).

    Thus, the derivative of \( 5 \ln t \) is \( \frac{5}{t} \).
This approach simplifies the process of working with logs. It breaks down the problem into manageable steps, allowing you to first handle the logarithmic differentiation and then adjust for any constants. Using logarithmic differentiation precisely and effectively is crucial, especially as you encounter more complex expressions.
Differentiation Techniques
Differentiation techniques refer to the various methods used to find the derivative of a function. Understanding when and how to apply each technique is essential.

In our exercise, two key techniques were used:
  • **Power Rule**: Used for polynomial terms like \( t^2 \). As simple as identifying the exponent \( n \) and applying the power rule to find the derivative.
  • **Logarithmic Differentiation**: Employed to tackle the \( 5 \ln t \) term. Here, the differentiation of a log function was facilitated using the constant rule.

    In any differentiation problem, the first step is to break down the function into simpler parts if possible. Different parts of a function might require different techniques.
  • Once you’ve identified which rule applies to which part, you apply it accordingly. Always combine the results from these parts at the end to get the complete derivative.
Differentiation techniques are like tools in a toolbox. Knowing which to pick for a particular job is crucial for efficiency and accuracy.

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

The distance, \(s,\) of a moving body from a fixed point is given as a function of time by \(s=20 e^{t / 2} .\) Find the velocity, \(v,\) of the body as a function of \(t\)

If \(p\) is price in dollars and \(q\) is quantity, demand for a product is given by $$q=5000 e^{-0.08 p}$$ (a) What quantity is sold at a price of \(\$ 10 ?\) (b) Find the derivative of demand with respect to price when the price is \(\$ 10\) and interpret your answer in terms of demand.

At a time \(t\) hours after it was administered, the concentration of a drug in the body is \(f(t)=27 e^{-0.14 t} \mathrm{ng} / \mathrm{ml}\). What is the concentration 4 hours after it was administered? At what rate is the concentration changing at that time?

The demand curve for a product is given by $$q=f(p)=10,000 e^{-0.25 p}$$ where \(q\) is the quantity sold and \(p\) is the price of the product, in dollars. Find \(f(2)\) and \(f^{\prime}(2) .\) Explain in economic terms what information each of these answers gives you.

If a someone is lost in the wilderness, the search and rescue team identifies the boundaries of the search area and then uses probabilities to help optimize the chances of finding the person, assuming the subject is immobile. The probability, \(O,\) of the person being outside the search area after the search has begun and the person has not been found, is given by $$O(E)=\frac{I}{1-(1-I) E}$$ where \(I\) is the probability of the person being outside the search area at the start of the search and \(E\) is the search effort, a measure of how well the search area has been covered by the resources in the field. (a) If there was a \(20 \%\) chance that the subject was not in the search area at the start of the search, and the search effort was \(80 \%,\) what is the current probability of the person being outside the search area? (Probabilities are between 0 and \(1,\) so \(20 \%=0.2\) and \(80 \%=0.8 .)\) (b) In practical terms, what does \(I=1\) mean? Is this realistic? (c) Evaluate \(O^{\prime}(E) .\) Is it positive or negative? What does that tell you about \(O\) as \(E\) increases?

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.