Problem 64 - Thomas Calculus

Question

In Exercises \(57-70\) , use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$ y=\frac{1}{t(t+1)(t+2)} $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{1}{t(t+1)(t+2)} \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right)\)."

Step 1: Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm of both sides of the equation to simplify the differentiation process. We have: \( \ln(y) = \ln \left( \frac{1}{t(t+1)(t+2)} \right) \). Apply the properties of logarithms to simplify: \[ \ln(y) = - \ln(t) - \ln(t+1) - \ln(t+2) \].

Step 2: Differentiate Both Sides With Respect to t

Differentiate both sides of the equation with respect to \(t\). Using the chain rule, the derivative of the left side is \( \frac{1}{y} \cdot \frac{dy}{dt} \). Differentiate the right side term by term: - The derivative of \(-\ln(t)\) is \(-\frac{1}{t}\).- The derivative of \(-\ln(t+1)\) is \(-\frac{1}{t+1}\).- The derivative of \(-\ln(t+2)\) is \(-\frac{1}{t+2}\).Combining these gives:\[ \frac{1}{y} \cdot \frac{dy}{dt} = -\frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \].

Step 3: Solve for \(\frac{dy}{dt}\)

We now solve for \(\frac{dy}{dt}\). Multiply both sides by \(y\) to isolate the derivative:\[ \frac{dy}{dt} = y \left( -\frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \right) \].

Step 4: Substitute Back the Expression for y

Replace \(y\) with the original function: \(y = \frac{1}{t(t+1)(t+2)}\). Thus,\[ \frac{dy}{dt} = \frac{1}{t(t+1)(t+2)} \left( -\frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \right) \].

Step 5: Simplify the Expression

Simplify the expression further by factoring out -1 from the terms in parentheses:\[ \frac{dy}{dt} = -\frac{1}{t(t+1)(t+2)} \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right) \]. This is the derivative of the given function.

Key Concepts

Understanding Derivatives

A derivative tells us how a function changes as its input changes. It's like a rate of change. Imagine you're driving and tracking how your speed changes over time. That's similar to how a derivative tracks the change of a function concerning its variable. Here's what it does:

• Measures the slope of a function at any given point.
• Helps understand how the function behaves—whether it's increasing, decreasing, or staying constant.

In calculus, to find the derivative of a function like the one in the exercise, we often use different techniques like logarithmic differentiation, especially when dealing with complex expressions. This specific method simplifies the process significantly when you have products or quotients inside a logarithm.

The Role of the Chain Rule

The chain rule is a powerful tool in calculus used whenever you want to differentiate a composite function. Think of it as a way to peel back layers of a mathematical onion.

• It helps take the derivative of the outer function, while respecting the inner function's influence.
• This rule is crucial when functions are nested, like when taking logarithms of other complicated functions.

In the exercise, after transforming the expression with a natural logarithm, we applied the chain rule to handle the differentiation of both sides of the equation. This allowed us to systematically find the slope of the function with respect to its variable, further aligning the terms appropriately for simplification.

Exploring Natural Logarithms

The natural logarithm, denoted as \(\ln\), is a special type of logarithm with a base of \(e\), where \(e\approx2.718\). It's widely used in calculus due to its nice properties:

• Allows simplification of multiplication into addition, making complex expressions easier to handle.
• Takes advantage of its derivative, which is simply \(\frac{1}{x}\).

In this particular exercise, taking the natural logarithm of both sides transforms a product into several addition terms. This transformation lends itself beautifully to the differentiation process and ultimately helps simplify our calculations, aligning the function for a more straightforward derivative finding process.

Importance of the Simplification Step

Simplification is an essential last step in many calculus problems. It makes the final expression more manageable and easier to interpret.

• Combines like terms to reduce complexity.
• Often requires factoring out constants or common factors.

In the original exercise, the final step of simplification involved factoring out \(-1\) from the expression, giving us a cleaner answer. A well-simplified expression not only looks better but is also easier to use in further calculations or practical applications. It helps ensure that the solution is as efficient and concise as possible, demonstrating a deeper understanding of mathematical manipulation.