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Chapter 3: Problem 78

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{3} r \cdot \log _{9} r$$

Short Answer

Expert verified
The derivative is \(\frac{dy}{dr} = \frac{1}{r}\left( \frac{\log_{9} r}{\ln 3} + \frac{\log_3 r}{\ln 9} \right)\).

Step by step solution

01

Break Down the Problem

The function is given as a product of two logarithms: \(y = \log_{3} r \cdot \log_{9} r\). To differentiate it, we'll use the product rule for derivatives, which states \((u \, v)' = u' \, v + u \, v'\), where \(u = \log_{3} r\) and \(v = \log_{9} r\).
02

Differentiate the First Logarithm

We need to find the derivative of \(u = \log_3 r\) with respect to \(r\). Using the change of base formula, \(\log_{3} r = \frac{\ln r}{\ln 3}\). Thus, \(u' = \frac{d}{dr}\left( \frac{\ln r}{\ln 3} \right) = \frac{1}{r \ln 3}\) since \(\ln 3\) is a constant.
03

Differentiate the Second Logarithm

Next, we find the derivative of \(v = \log_9 r\), using the change of base formula: \(\log_{9} r = \frac{\ln r}{\ln 9}\). Therefore, \(v' = \frac{d}{dr}\left( \frac{\ln r}{\ln 9} \right) = \frac{1}{r \ln 9}\), since \(\ln 9\) is a constant.
04

Apply the Product Rule

Apply the product rule for differentiation to find \(\frac{dy}{dr}\): \[\frac{dy}{dr} = \left( \frac{1}{r \ln 3} \right) \log_9 r + \log_3 r \cdot \left( \frac{1}{r \ln 9} \right) \]This expression reflects the derivative based on the product rule and our earlier calculations.
05

Simplify the Derivative

Combine the terms over a common denominator: \[\frac{dy}{dr} = \frac{1}{r}\left( \frac{\log_{9} r}{\ln 3} + \frac{\log_3 r}{\ln 9} \right)\]The expression is fully simplified with a single fraction.

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Key Concepts

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

product rule
The product rule is a fundamental concept in calculus differentiation that helps us find the derivative of a product of two functions. If we have a function expressed as the product of two separate functions, such as \(y = u \, v\), where \(u\) and \(v\) are both functions of the same variable, the product rule provides a systematic way to differentiate this product.To apply product rule, the formula we use is \[(u \, v)' = u' \, v + u \, v'\]. This means, to differentiate the product of \(u\) and \(v\), we first differentiate \(u\) while keeping \(v\) the same and add it to \(u\) times the derivative of \(v\).
  • Step 1: Differentiate \(u\), leaving \(v\) unchanged.
  • Step 2: Differentiate \(v\), leaving \(u\) unchanged.
  • Step 3: Multiply each derivative by the unchanged counterpart and sum them up.
In our exercise, \(u\) was set as \(\log_{3} r\) and \(v\) as \(\log_{9} r\). By understanding and applying these steps, you can effectively use the product rule in similar problems, making it easier to navigate through differentiation tasks involving products.
logarithmic differentiation
Logarithmic differentiation is a useful technique particularly when dealing with products or quotients of functions, or functions raised to a power. It simplifies the process of differentiation by leveraging the properties of logarithms. The process involves the following:
  • Taking the natural logarithm of both sides of a potentially complex function.
  • Applying the rules of logarithms to simplify the expression.
  • Differentiating the simplified expression with respect to the variable of interest.
By utilizing logarithmic differentiation, computations become more manageable, especially for functions involving products or complicated exponents. Logarithmic differentiation can also help in cases where directly applying standard differentiation rules might be cumbersome or complicated. In our problem, although explicit logarithmic differentiation wasn't applied to the entire expression, knowledge of logarithms assisted in differentiating the given logarithmic terms using their derivatives.
change of base formula
The change of base formula is a widely-used formula in mathematics that helps in converting logarithms from one base to another, particularly converting to natural logarithms which makes differentiation straightforward.The formula is given as:\[\log_{b} a = \frac{\ln a}{\ln b}\]This formula is extremely useful because the natural logarithm \(\ln\) is well-integrated into calculus due to its properties, particularly its derivative.In our exercise, the change of base formula was essential to convert both \(\log_{3} r\) and \(\log_{9} r\) into forms involving natural logarithms:
  • \(\log_{3} r = \frac{\ln r}{\ln 3}\)
  • \(\log_{9} r = \frac{\ln r}{\ln 9}\)
By doing so, it allowed us to easily find the derivatives as the natural logarithm's derivative is simply \(\frac{1}{r}\). This process is a key step whenever differentiating expressions involving complex logarithms, as it simplifies calculations by using the more ubiquitous natural logarithm base.

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Most popular questions from this chapter

Use your graphing utility. Graph the rational function \(y=\left(2-x^{2}\right) / x^{2} .\) Then graph \(y=\) \(\cos \left(2 \sec ^{-1} x\right)\) in the same graphing window. What do you see? Explain.

Find the derivative of \(y\) with respect to the given independent variable. $$y=t \log _{3}\left(e^{(\sin t)(\ln 3)}\right)$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=x^{\pi}$$

a. Find the linearization of \(f(x)=2^{x}\) at \(x=0 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together for \(-3 \leq x \leq 3\) and \(-1 \leq x \leq 1\)

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 the 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 \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.
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