/*! 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 24 Find the derivative of \(y\) wit... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the derivative of \(y\) with respect to the given independent variable. \(y=\log _{3}(1+\theta \ln 3)\)

Short Answer

Expert verified
The derivative is \( \frac{1}{1+\theta \ln 3} \).

Step by step solution

01

Understand the Function

We have the function given as \( y = \log_{3}(1+\theta \ln 3) \). This is a logarithmic function with base 3. The expression inside the logarithm is \( 1+\theta \ln 3 \).
02

Convert Logarithmic Base

Use the change of base formula for logarithms: \( \log_{a}(b) = \frac{\ln(b)}{\ln(a)} \). By applying this conversion to our function, we have: \[ y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \].
03

Differentiate with Respect to \( \theta \)

Differentiate \( y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \) with respect to \( \theta \):Using the chain rule: \[ \frac{dy}{d\theta} = \frac{1}{\ln 3} \cdot \frac{d}{d\theta}(\ln(1+\theta \ln 3)) \].
04

Differentiate the Natural Logarithm

The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{d\theta} \). Let \( u = 1 + \theta \ln 3 \), then \( \frac{du}{d\theta} = \ln 3 \). Thus, the derivative of the natural logarithm is:\[ \frac{d}{d\theta}(\ln(1+\theta \ln 3)) = \frac{1}{1+\theta \ln 3} \cdot \ln 3 \].
05

Finalize the Derivative

Substitute the result from Step 4 back into the expression from Step 3:\[ \frac{dy}{d\theta} = \frac{1}{\ln 3} \cdot \frac{\ln 3}{1+\theta \ln 3} = \frac{1}{1+\theta \ln 3} \]. This is the derivative of \( y \) with respect to \( \theta \).

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Key Concepts

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

Logarithmic Function
Logarithmic functions are an essential part of calculus and algebra. These functions help us solve problems related to exponential growth and decay by providing an inverse operation to exponentiation. In the given exercise, we deal with a logarithmic function in a specified base, which is base 3. The function is written as \( y = \log_{3}(1+\theta \ln 3) \). This notation shows that the function takes \(1 + \theta \ln 3\) as an argument and calculates how many times 3 must be raised to get that result.

One common challenge with logarithmic functions is understanding the base. Many students often encounter logarithms in base 10 or as natural logarithms with base \(e\). In this exercise, the base 3 logarithm needs conversion to a more familiar base using a special formula, called the change of base formula. Understanding the roles of different bases and how to manage them is crucial when learning about logarithmic functions.
Chain Rule
In calculus, the chain rule is a fundamental tool for finding the derivative of composite functions. A composite function is a function made up of other functions nested within it, such as \( f(g(x)) \). In this exercise, our function is a composition involving the natural logarithm, \( \ln(1 + \theta \ln 3) \), nested inside a division by a constant, \( \ln 3 \).When we differentiate composite functions, the chain rule allows us to "peel back" these layers. The rule requires taking the derivative of the outer function and multiplying it by the derivative of the inside function. It's like a sequence of multiplying changes. For example:
  • Take the derivative of \( y = \ln u / \ln 3 \).
  • Identify \( u = 1 + \theta \ln 3 \) and find its derivative with respect to \( \theta \).
This results in a beautiful orchestration of derivatives that simplifies complex differentiation processes, helping us tackle the exercise effectively.
Change of Base Formula
The change of base formula is an incredibly handy mathematical rule that allows us to convert logarithms from one base to another. This is particularly beneficial when the base of a logarithm does not match common or natural bases (like 10 or \(e\)) we often use in calculus.In this exercise, we start with a logarithmic function in base 3. Due to the base, direct differentiation can pose challenges. The change of base formula helps us rewrite \( y = \log_{3}(1+\theta \ln 3) \) as \( y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \). This formula:
  • \( \log_{a}(b) = \frac{\ln(b)}{\ln(a)} \)
allows us to switch logarithmic bases efficiently, making differentiation or solving equations much more straightforward. By transforming to natural logarithms, we engage tools like the chain rule seamlessly and smooth the way to our derivative solution.

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Most popular questions from this chapter

Volume A region in the first quadrant is bounded above by the curve \(y=\cosh x,\) below by the curve \(y=\sinh x,\) and on the left and right by the \(y\) -axis and the line \(x=2,\) respectively. Find the volume of the solid generated by revolving the region about the \(x\) -axis.

When hyperbolic function keys are not available on a calculator, it is still possible to evaluate the inverse hyperbolic functions by expressing them as logarithms, as shown here. $$ \begin{array}{rlrl}{\sinh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {-\infty < x < \infty} \\ {\cosh ^{-1} x} & {=\ln \left(x+\sqrt{x^{2}+1}\right),} & {x \geq 1} \\ {\tanh ^{-1} x} & {=\frac{1}{2} \ln \frac{1+x}{1-x},} & {|x| < 1} \\ {\operatorname{sech}^{-1} x} & {=\ln \left(\frac{1+\sqrt{1-x^{2}}}{x}\right),} & {0 < x \leq 1} \\\ {\operatorname{csch}^{-1} x} & {=\ln \left(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|}\right),} & {x \neq 0} \\\ {\operatorname{coth}^{-1} x} & {=\frac{1}{2} \ln \frac{x+1}{x-1},} & {|x| > 1}\end{array} $$ Use the formulas given above to express the numbers in Exercises \(61-66\) in terms of natural logarithms. $$ \tanh ^{-1}(-1 / 2) $$

Evaluate the integrals in Exercises \(95-104\) $$ \int_{1}^{2} \frac{8 d x}{x^{2}-2 x+2} $$

The region between the curve \(y=\sec ^{-1} x\) and the \(x\) -axis from \(x=1\) to \(x=2\) (shown here) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid.

a. Find the centroid of the curve \(y=\cosh x,-\ln 2 \leq x \leq \ln 2\) b. Evaluate the coordinates to two decimal places. Then sketch the curve and plot the centroid to show its relation to the curve.

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