Problem 62 - Thomas Calculus in SI Units

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

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

Step 1: Take the Natural Logarithm of Both Sides

To start logarithmic differentiation, we take the natural logarithm of both sides of the equation. This gives us: \[ \ln(y) = \ln \left( \frac{1}{t(t+1)(t+2)} \right) \] Using logarithm rules, we simplify the right side: \[ \ln(y) = - \ln(t) - \ln(t+1) - \ln(t+2) \]

Step 2: Differentiate Both Sides with Respect to t

Next, differentiate both sides of the equation with respect to \(t\). The derivative of \(\ln(y)\) with respect to \(t\) requires the chain rule:\[ \frac{d}{dt} \ln(y) = \frac{1}{y} \cdot \frac{dy}{dt} \]On the right side, differentiate each term:\[ - \frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \]

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

Set the derivatives equal to each other from Step 2:\[ \frac{1}{y} \cdot \frac{dy}{dt} = - \frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \]Solve for \(\frac{dy}{dt}\) by multiplying both sides by \(y\):\[ \frac{dy}{dt} = y \left( - \frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \right) \]

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

Remember, \( y = \frac{1}{t(t+1)(t+2)} \). Substitute this back into the expression for \(\frac{dy}{dt}\):\[ \frac{dy}{dt} = \frac{1}{t(t+1)(t+2)} \left( - \frac{1}{t} - \frac{1}{t+1} - \frac{1}{t+2} \right) \] This completes the derivative of \(y\) with respect to \(t\).

Key Concepts

Natural Logarithm

The natural logarithm is a special type of logarithm where the base is the number \(e\) (approximately 2.71828). It's an essential tool in calculus when differentiating complex functions, especially those involving products, quotients, or powers.
To compute the natural logarithm of a function, we write \(\ln(y)\), which stands for the natural logarithm of \(y\). By applying the natural logarithm to both sides of an equation, we can simplify complex products or quotients into sums or differences, making them easier to differentiate.
For instance, take a product \(x^ay^b\). Using logarithmic properties, we simplify to:
- \(\ln(x^ay^b) = a\ln(x) + b\ln(y)\)
This transforms multiplication into addition. This relationship is crucial for differentiating products and quotients in functions like \(y = \frac{1}{t(t+1)(t+2)}\).

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. When dealing with a function of a function, the chain rule helps us differentiate the outer function and multiply it by the derivative of the inner function.

Consider the function \(f(g(x))\). The chain rule tells us that the derivative \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\).
In logarithmic differentiation, when we differentiate \(\ln(y)\) with respect to \(t\), the chain rule is employed. The derivative of \(\ln(y)\) is \(\frac{1}{y}\), and then we multiply by \(\frac{dy}{dt}\), giving us \(\frac{d}{dt} \ln(y) = \frac{1}{y} \cdot \frac{dy}{dt}\). Using the chain rule ensures we correctly account for the relationship between \(y\) and \(t\).

Product Rule

The product rule is useful when differentiating functions that are multiplied together. It states that the derivative of a product of two functions \(u(t)\) and \(v(t)\) is \(u'(t)v(t) + u(t)v'(t)\).
This is especially important when handling multiple terms that are products, like in polynomial expressions.
Although not directly applied in the solution to this problem, the simplification of expressions into sums via logarithms in logarithmic differentiation often results in an easier path for differentiation using the chain and product rules.
By decomposing complex products and applying derivatives to each term separately, we ease the calculation process, essentially transforming a complicated task into a series of simpler operations.

Derivative of a Quotient

The derivative of a quotient follows the quotient rule in calculus. When dealing with functions that are divided by one another, the quotient rule becomes crucial.
For a quotient \(\frac{f(t)}{g(t)}\), the derivative is found using: \[\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}\]In the current problem, logarithmic differentiation allows us to avoid directly applying the quotient rule by transforming the quotient into a sum using logarithms:

• The right side of \(\ln(y) = \ln \left( \frac{1}{t(t+1)(t+2)} \right)\) simplifies to
  \(-\ln(t) - \ln(t+1) - \ln(t+2)\)

This eliminates the need for the complex application of the quotient rule. It simplifies the differentiation process by handling simple sums instead, allowing for easier calculation of the derivative.