Problem 76 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative of \(y\) is \(\frac{2 \ln r}{r \ln 3 \cdot \ln 9}\).

Step 1: Simplify the Expression

Notice that the function is a product of two logarithms with different bases: \(y = \log_{3} r \cdot \log_{9} r\). We can use the change of base formula to convert the logarithms to the same base. Recall the change of base formula: \(\log_{a} b = \frac{\ln b}{\ln a}\). Let's convert both logarithms to natural logarithms: \(\log_{3} r = \frac{\ln r}{\ln 3}\) and \(\log_{9} r = \frac{\ln r}{\ln 9}\). The function then becomes: \[y = \frac{\ln r}{\ln 3} \cdot \frac{\ln r}{\ln 9} = \frac{(\ln r)^2}{\ln 3 \cdot \ln 9}.\]

Step 2: Differentiate the Simplified Expression

With the expression simplified to \(y = \frac{(\ln r)^2}{\ln 3 \cdot \ln 9}\), we now need to find \(\frac{dy}{dr}\). Notice it's a simple function of the form \(C \cdot u^2\) where \(C = \frac{1}{\ln 3 \cdot \ln 9}\) is a constant and \(u = \ln r\). We use the power rule and the chain rule:\[ \frac{dy}{dr} = C \cdot 2u \cdot \frac{du}{dr} = \frac{2 \ln r}{\ln 3 \cdot \ln 9} \cdot \frac{1}{r} = \frac{2 \ln r}{r \ln 3 \cdot \ln 9}. \]

Step 3: Simplified Derivative

The derivative from step 2 is: \(\frac{dy}{dr} = \frac{2 \ln r}{r \ln 3 \cdot \ln 9}\). This is the final expression for the derivative of \(y\) with respect to \(r\).

Key Concepts

Logarithms

Logarithms are used to find the power to which a base number must be raised to produce a given number. For example, the logarithm base 10 of 100 is 2, because 10 to the power of 2 is 100. In mathematical notation, if we say \( \log_{b}(a) = c \), it means \( b^{c} = a \).

• The logarithm is helpful in dealing with exponential growth or decay, such as in finance, certain scientific phenomena, and sound intensity.
• Changing the base of logarithms is a common technique when dealing with derivatives, as seen in this problem.
• Natural logarithms use the base \( e \), a mathematical constant approximately equal to 2.71828.

Understanding how to manipulate logarithms makes complex expressions more manageable. You often convert them to the natural logarithm due to its convenient properties for differentiation.

Chain Rule

The chain rule is a formula in calculus used to differentiate composite functions. Composite functions are where one function is nested inside another function, like \( h(x) = f(g(x)) \). The chain rule helps break down these layers.

• According to the chain rule, if you have a function \( y = f(g(x)) \), the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This rule allows the differentiation to proceed step-by-step: differentiate the outer function first while keeping the inner function intact, and then multiply it by the derivative of the inner function.
• In our problem, we apply the chain rule to logarithmic functions, which often appear as composite due to their argument.

The chain rule is very versatile and powerful, especially when functions are deeply nested. It's almost like peeling an onion layer by layer!

Change of Base Formula

The change of base formula is crucial when dealing with logarithms of different bases. It allows you to rewrite logarithms in terms of more convenient bases like natural logarithms.

• The formula is \( \log_{a} b = \frac{\ln b}{\ln a} \). This reformulates logarithms to a base that you prefer, often base \(e\) or 10.
• In this exercise, we used the change of base formula to convert \( \log_3 r \) and \( \log_9 r \) into natural logs, simplifying our work during differentiation.
• This is particularly useful in calculus, as natural logarithms have well-known derivatives.

By employing this formula, we made a seemingly complex differentiation problem much easier to manage and solve.

Power Rule

The power rule is one of the fundamental rules for finding the derivative of polynomial expressions and is widely used in calculus.

• For a function \( y = x^n \), the power rule states that the derivative \( \frac{dy}{dx} \) is \( n \cdot x^{(n-1)} \).
• In the context of this problem, even though we deal with logarithms, we ended up using the power rule because our final expression had the form \( C \cdot u^2 \), where \( C \) is a constant and \( u \) contains a logarithmic function.
• By applying the power rule along with the chain rule, we systematically found the derivative.

The power rule simplifies the process of taking derivatives, especially in polynomial forms, by straightforwardly reducing the power of the function's terms.