/*! 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 Find the derivative of the funct... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the derivative of the functions. $$y=5+\ln (3 t+2)$$

Short Answer

Expert verified
The derivative is \(\frac{3}{3t+2}\).

Step by step solution

01

Differentiate Constant Term

In the given function \(y = 5 + \ln(3t + 2)\), notice that the term \(5\) is a constant. The derivative of any constant term is \(0\). Thus, we do not need to differentiate this part further.
02

Apply the Derivative of Natural Logarithm

For the natural logarithmic function \(\ln(3t + 2)\), we use the formula for the derivative of \(\ln u\), which is \(\frac{1}{u} \cdot \frac{du}{dt}\). Here, \(u = 3t + 2\), so we need to find the derivative of \(3t + 2\).
03

Differentiate the Inside Function

Find the derivative of \(u = 3t + 2\) with respect to \(t\). The derivative is \(\frac{du}{dt} = 3\) because the derivative of \(3t\) is \(3\) and the derivative of the constant \(2\) is \(0\).
04

Combine Results to Find Derivative

Substitute \(\frac{du}{dt} = 3\) into the derivative formula for the logarithmic function. The derivative of \(\ln(3t + 2)\) is \(\frac{1}{3t + 2} \cdot 3 = \frac{3}{3t + 2}\).
05

Write the Final Derivative

Combine the results of all parts. Since the derivative of the constant \(5\) is \(0\), and the derivative of \(\ln(3t + 2)\) is \(\frac{3}{3t + 2}\), the final derivative of the function \(y = 5 + \ln(3t + 2)\) is \(\frac{3}{3t + 2}\).

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Key Concepts

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

Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. This special logarithm is commonly used in mathematics due to its unique properties which simplify many calculus operations. One significant property of the natural logarithm is its derivative. If you have a function \( \ln(u) \), where \( u \) is a function of \( t \), its derivative is \( \frac{1}{u} \times \frac{du}{dt} \). This formula allows us to find the rate at which the natural logarithm function changes with respect to time or any variable \( t \). Natural logarithms appear frequently in scientific calculations and help model exponential growth processes like population growth and radioactive decay. Understanding the derivative of \( \ln \) forms the basis for solving complex calculus problems involving logarithmic differentiation.
Differentiation
Differentiation is the process of finding the derivative of a function. In simple terms, a derivative represents the rate at which a function changes at any given point. It's equivalent to finding the slope of the tangent line to the function at a particular point. For example, consider the function \( y = f(t) \). The derivative \( f'(t) \) of \( y \) is determined by how \( y \) changes as \( t \) slightly increases. Differentiation can apply to constants, algebraic expressions, exponential functions, and more. In the given problem, we start with a simple differentiation of the constant term \( 5 \). The derivative of any constant is always zero because constants do not change and thus have no rate of change. Understanding how to differentiate each part of a composite function, like \( 5 + \ln(3t + 2) \), allows us to combine these parts successfully for a complete derivative.
Chain Rule
The Chain Rule is a fundamental theorem in calculus that helps differentiate composite functions. Composite functions are combinations of two or more functions, where one function is nested within another, such as \( \ln(3t + 2) \). The Chain Rule states: if you have two functions \( f(g(x)) \), then the derivative \( f'(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Essentially, you differentiate the outer function and multiply by the derivative of the inner function. When applying the Chain Rule in the example problem, we treat \( u = 3t + 2 \) as the inner function and \( \ln(u) \) as the outer function. First, we differentiate the inner function \( u = 3t + 2 \) to get \( \frac{du}{dt} = 3 \). Then, we apply it to the derivative formula for \( \ln(u) \), resulting in the outer derivative being \( \frac{1}{3t + 2} \) multiplied by the inner \( \frac{du}{dt} \) which is 3, giving \( \frac{3}{3t + 2} \). The utilization of the Chain Rule simplifies working with nested functions and is crucial for accurate differentiation of complex expressions.

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Most popular questions from this chapter

Differentiate the functions in Problems. Assume that \(A\) \(B,\) and \(C\) are constants. $$R=3 \ln q$$

(a) Find the equation of the tangent line to \(y=\ln x\) at \(x=1.\) (b) Use it to calculate approximate values for \(\ln (1.1)\) and \(\ln (2).\) (c) Using a graph, explain whether the approximate values are smaller or larger than the true values. Would the same result have held if you had used the tangent line to estimate \(\ln (0.9)\) and \(\ln (0.5) ?\) Why?

A ball is dropped from the top of the Empire State Building. The height, \(y,\) of the ball above the ground (in feet) is given as a function of time, \(t,\) (in seconds) by $$y=1250-16 t^{2}$$ (a) Find the velocity of the ball at time \(t .\) What is the sign of the velocity? Why is this to be expected? (b) When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour \((1 \mathrm{ft} / \mathrm{sec}=15 / 22 \mathrm{mph})\)

Find the equation of the tangent line to \(f(x)=10 e^{-0.2 x}\) at \(x=4.\)

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?

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