Problem 86 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=3 \log _{8}\left(\log _{2} t\right)$$

Step-by-Step Solution

Answer

\(\frac{dy}{dt} = \frac{3}{t \ln 8 \cdot \ln 2 \cdot \log_2 t}\).

Step 1: Recall the Chain Rule

The function given is a composition of functions, so we will need to use the chain rule. If you have a function of the form \(y = f(g(x))\), then the derivative \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). In our problem, \(y = 3 \log_8(\log_2 t)\), so this contains a logarithmic function within another logarithmic function.

Step 2: Differentiate the Outer Function

Consider the outer function \(u = 3 \log_8(v)\) where \(v = \log_2 t\). The derivative of \(\log_a x\) is \(\frac{1}{x \ln a}\). Therefore, \(\frac{du}{dv} = 3 \cdot \frac{1}{v \ln 8}\).

Step 3: Differentiate the Inner Function

Now differentiate the inner function \(v = \log_2 t\). The derivative of \(\log_a x\) is \(\frac{1}{x \ln a}\), so \(\frac{dv}{dt} = \frac{1}{t \ln 2}\).

Step 4: Apply the Chain Rule

By the chain rule, the derivative of the original function \(y = 3 \log_8(\log_2 t)\) with respect to \(t\) is found by multiplying the derivative of the outer function by the derivative of the inner function: \[\frac{dy}{dt} = \frac{du}{dv} \cdot \frac{dv}{dt} = \left(3 \cdot \frac{1}{v \ln 8}\right) \cdot \left(\frac{1}{t \ln 2}\right).\]

Step 5: Substitute and Simplify

We now substitute \(v = \log_2 t\) into the equation. This gives: \[\frac{dy}{dt} = \frac{3}{(\log_2 t) \cdot t \ln 8 \cdot \ln 2}.\] By simple algebra, this can also be written as \[\frac{dy}{dt} = \frac{3}{t \ln 8 \cdot \ln 2 \cdot \log_2 t}.\]

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus that helps in finding the derivative of composite functions. In simple terms, a composite function is when you have one function inside another. For example, in our exercise, we have two logarithms nested within each other. This creates a situation where we cannot directly find the derivative.

To use the chain rule, you follow these steps:

• Identify the inner and outer functions. In the given example, the outer function is the logarithm base 8, and the inner function is the logarithm base 2.
• Differentiate the outer function as if the inner function were a simple variable.
• Multiply the result by the derivative of the inner function. This step completes the application of the chain rule.

This method succinctly allows us to differentiate complex compositions by breaking them down into manageable parts.

Logarithmic Differentiation

Logarithmic differentiation comes in handy when working with functions that involve multiple logarithms or exponents. It uses the properties of logarithms to simplify differentiation. In our context, it helps with functions like \(\log_a x\), where taking the derivative directly might be challenging.

Here's how logarithmic differentiation was applied:

• We first took the derivative of the outermost logarithmic function. For a function \( \log_a x \), the derivative is \( \frac{1}{x \ln a} \).
• The function effectively becomes manageable by leveraging logarithm rules, making it easier to handle complex real-world problems.

This approach is useful, not only for simplifying derivatives but also for gaining insights into the behavior of the function by examining the rate of change.

Composition of Functions

The concept of composition of functions is central to understanding how to apply the chain rule and logarithmic differentiation. When you compose functions, you essentially plug one function into another. This allows complex functions to be expressed in terms of simpler functions.

In our case:

• The function \( y = 3 \log_8(\log_2 t) \) is a composition where \( \log_2 t \) is inside \( 3 \log_8(x) \).
• Recognizing this lets us smartly decompose the problem into smaller, easily solvable parts using the chain rule.
• By differentiating each part separately and then combining the results, we efficiently handle the complexity involved.

Understanding composition not only aids in differentiation but also plays a vital role in other mathematical operations where functions are involved.