Problem 48 - University Calculus: Early Transcendentals

Question

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

\( \frac{dy}{dt} = -\frac{1}{t^2(t+1)(t+2)} - \frac{1}{t(t+1)^2(t+2)} - \frac{1}{t(t+1)(t+2)^2} \).

Step 1: Express the Function in a Logarithmic Form

Given the function \( y = \frac{1}{t(t+1)(t+2)} \), we take the natural logarithm of both sides to simplify the expression for differentiation. This gives us: \[ \ln y = \ln \left( \frac{1}{t(t+1)(t+2)} \right) = -\ln(t) - \ln(t+1) - \ln(t+2). \]

Step 2: Differentiate Both Sides with Respect to \( t \)

Differentiating the left side gives us \( \frac{1}{y} \frac{dy}{dt} \). Differentiating the right side involves applying the chain rule to each 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 \( -\frac{1}{t+2} \).Thus, we have: \[ \frac{1}{y} \frac{dy}{dt} = -\frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2}. \]

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

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

Step 4: Substitute \( y \) Back into the Expression

Since \( y = \frac{1}{t(t+1)(t+2)} \), substitute this back into the expression for \( \frac{dy}{dt} \):\[ \frac{dy}{dt} = \frac{1}{t(t+1)(t+2)} \left( -\frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \right). \]

Step 5: Simplify the Expression

Simplify the expression by multiplying the terms:\[ \frac{dy}{dt} = -\frac{1}{t^2(t+1)(t+2)} - \frac{1}{t(t+1)^2(t+2)} - \frac{1}{t(t+1)(t+2)^2}. \]Combine terms if possible and simplify further if needed.

Key Concepts

Derivative Calculation

Understanding how to calculate derivatives is crucial when dealing with complex functions, especially those that can be simplified using logarithmic techniques. In the given exercise, we aim to find the derivative of a function expressed as a fraction of polynomial terms: \[ y = \frac{1}{t(t+1)(t+2)} \] To simplify the differentiation process, we take the natural logarithm of both sides, yielding a sum of simpler logarithmic terms. This conversion is a strategic move because taking derivatives of logarithmic expressions can be more straightforward, especially when combined with the properties of logarithms.

• Firstly, transform the product inside the fraction into a sum of logs: \[ \ln y = -\ln(t) - \ln(t+1) - \ln(t+2) \]
• Next, we differentiate both sides with respect to the independent variable, \( t \).
• The left side gives us the derivative expression: \[ \frac{1}{y} \frac{dy}{dt} \]

This established framework is a hallmark of the method known as **Logarithmic Differentiation**. It cleverly simplifies complex multiplicative components by leveraging the properties of logarithms, allowing for a more manageable differentiation process.

Natural Logarithm

The **Natural Logarithm** plays a pivotal role in simplifying our original function. Its unique properties make it a powerful tool for differentiation. The natural logarithm, denoted as \( \ln \), is particularly handy because its derivative is straightforward, \( \frac{d}{dt} \ln(t) = \frac{1}{t} \).Using natural logarithms, we transform a complex product into a more approachable sum:

• This transformation involves changing the multiplicative terms \( t(t+1)(t+2) \) into additive terms, using the identity: \[ \ln \Big(\frac{1}{t(t+1)(t+2)} \Big) = -\ln(t) - \ln(t+1) - \ln(t+2) \]
• The minus signs come from the reciprocal in the original function \( y = \frac{1}{t(t+1)(t+2)} \).

By reducing the complexity with logarithms, you transform the original function in a manner conducive to easier **derivative calculations**. Hence, it's not just any logarithm but the natural log's precision and simplicity that enable a seamless differentiation process.

Chain Rule

The **Chain Rule** is indispensable in calculus, especially when differentiating complex functions. It serves as the bridge for the derivatives of composite functions, allowing us to handle each component of the function separately and efficiently.In this exercise, once we log-transform the function, we apply the chain rule to each natural log term individually:

• For \( -\ln(t) \), the derivative is: \[ -\frac{1}{t} \]
• For \( -\ln(t+1) \), the derivative becomes: \[ -\frac{1}{t+1} \]
• For \( -\ln(t+2) \), the differentiation results in: \[ -\frac{1}{t+2} \]

By applying the chain rule to these separate components after logarithmic transformation, we obtain derivatives that are perfectly aligned with the structure of the given function. The beauty of logarithmic differentiation is highlighted as the chain rule facilitates seamless and efficient computation of derivatives for each term.