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Chapter 3: Problem 74

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 \( \frac{dy}{d\theta} = \frac{1}{1 + \theta \ln 3} \).

Step by step solution

01

Understand the Function

The function given is \( y = \log_3(1 + \theta \ln 3) \). It is expressed using the logarithm with base 3. Understanding this function is key before taking any derivatives.
02

Convert the Logarithm Base

To differentiate logarithmic functions where the base is not \(e\), it is beneficial to convert the logarithm from base 3 to a natural logarithm. Use the change of base formula:\[y = \frac{\ln (1 + \theta \ln 3)}{\ln 3}\]Now the function is expressed in terms of natural logarithms, which we can differentiate more easily.
03

Apply the Derivative Rules

Differentiate the function \( y = \frac{\ln (1 + \theta \ln 3)}{\ln 3} \) with respect to \( \theta \). Notice that \( \ln 3 \) is a constant:\[\frac{dy}{d\theta} = \frac{1}{\ln 3} \cdot \frac{d}{d\theta}[\ln (1 + \theta \ln 3)]\]
04

Find the Inner Derivative

Find the derivative of the inner function using the chain rule. The derivative of \( \ln (1 + \theta \ln 3) \) is calculated as follows:\[\frac{d}{d\theta} [\ln (1 + \theta \ln 3)] = \frac{1}{1 + \theta \ln 3} \cdot \frac{d}{d\theta}(1 + \theta \ln 3)\]
05

Differentiate the Inner Expression

Now differentiate the expression \( 1 + \theta \ln 3 \) with respect to \( \theta \):\[\frac{d}{d\theta} (1 + \theta \ln 3) = \ln 3\]
06

Combine Results

Combine the results from Step 4 and Step 5 to find \( \frac{d}{d\theta} \ln (1 + \theta \ln 3) \):\[\frac{d}{d\theta} [\ln (1 + \theta \ln 3)] = \frac{\ln 3}{1 + \theta \ln 3}\]
07

Finish the Derivative Calculation

Substitute the result from Step 6 back into the expression found in 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.

Chain Rule
The chain rule is an essential tool in calculus for differentiating composite functions. It allows us to find the derivative of a complex expression by breaking it down into simpler parts. Imagine you have a function nested within another function, like peeling an onion. This is where the chain rule shines.

The chain rule formula is:
  • If you have a function expressed as \( f(g(x)) \), the derivative \( \frac{d}{dx}f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
In the given exercise, the chain rule is applied when differentiating \( \ln(1 + \theta \ln 3) \). Here, the outer function is the natural logarithm \( \ln(u) \) and the inner function is \( u = 1 + \theta \ln 3 \). The outer derivative \( \frac{1}{u} \) is then multiplied by the derivative of the inner function, which is \( \ln 3 \).

Understanding the chain rule allows you to neatly handle these nested functions. It provides a framework to split up derivatives into manageable parts.
Logarithmic Differentiation
Logarithmic differentiation is a very useful technique when dealing with complicated products or quotients of functions, especially those involving powers and logarithms. This method often simplifies the process significantly.

In logarithmic differentiation, you leverage the property of logarithms where the log of a product becomes the sum of logs, making the differentiation process more straightforward:
  • Start by taking the natural logarithm of both sides of the equation.
  • Differentiate implicitly with respect to the independent variable.
In the exercise, although direct logarithmic differentiation is not explicitly used, understanding this concept is beneficial. Converting the base of the logarithm from 3 to \( e \) using the change of base formula is a key stepping stone.

This move helps in applying the standard rules of differentiation for natural logarithms, illustrating how logs can simplify complex expressions.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is a fundamental constant in mathematics that appears in various growth, decay, and mathematical modeling scenarios.

One of the appealing features of natural logarithms is their differentiation rule. The derivative of \( \ln(x) \) is simply \( \frac{1}{x} \), which simplifies many calculus problems. In practice, this makes the natural logarithm especially handy for calculus operations and is often the reason why we convert other logarithmic bases to base \( e \).

In the exercise, the original problem involved a logarithm with a base of 3. By converting it to a natural logarithm using the formula \( \log_3(x) = \frac{\ln(x)}{\ln(3)} \), we simplify our differentiation process. This change isn't just a cosmetic alteration鈥攊t's aligning the problem with the techniques most efficient in calculus, allowing easier application of the derivative rules.

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Most popular questions from this chapter

Use a CAS to perform the following steps. a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$2 y^{2}+(x y)^{1 / 3}=x^{2}+2, \quad P(1,1)$$

a. About how accurately must the interior diameter of a 10-m-high cylindrical storage tank be measured to calculate the tank's volume to within \(1 \%\) of its true value? b. About how accurately must the tank's exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within \(5 \%\) of the true amount?

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(3 \theta e^{-\theta}\right)$$

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(\frac{e^{\theta}}{1+e^{\theta}}\right)$$

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on Earth's surface, depending on the change in \(g\). By keeping track of \(\Delta T\), we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\) a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001\) sec. Find \(d g\) and estimate the value of g at the new location.
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