Problem 61 - Thomas Calculus in SI Units

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=t(t+1)(t+2)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = 3t^2 + 6t + 2 \).

Step 1: Express the Function in Logarithmic Form

Given the function \(y = t(t+1)(t+2)\), take the natural logarithm of both sides to facilitate differentiation. This gives us:\[ \ln y = \ln (t(t+1)(t+2)) \]. Use the logarithmic property: \(\ln(abc) = \ln a + \ln b + \ln c\), resulting in:\[ \ln y = \ln t + \ln(t+1) + \ln(t+2) \].

Step 2: Differentiate Both Sides with Respect to t

Differentiate both sides of the equation \(\ln y = \ln t + \ln(t+1) + \ln(t+2)\) with respect to \(t\). For the left side, use the derivative rule for \(\ln y\): \(\frac{1}{y}\frac{dy}{dt}\). For the right side, apply the derivative of \(\ln t\), \(\ln(t+1)\), and \(\ln(t+2)\), which are \(\frac{1}{t}\), \(\frac{1}{t+1}\), and \(\frac{1}{t+2}\) respectively. This yields:\[\frac{1}{y}\frac{dy}{dt} = \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2}\].

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

Multiply both sides of the equation from Step 2 by \(y\) to solve for \(\frac{dy}{dt}\):\[\frac{dy}{dt} = y \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right)\].

Step 4: Substitute \(y\) Back in Terms of \(t\)

Substitute the original function back for \(y\) in \(\frac{dy}{dt}\):\[\frac{dy}{dt} = t(t+1)(t+2) \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right)\]. Expand the expression to obtain\[\frac{dy}{dt} = (t+1)(t+2) + t(t+2) + t(t+1)\]. Simplifying this yields\[\frac{dy}{dt} = 3t^2 + 6t + 2\].

Key Concepts

Differentiation Techniques

Differentiation techniques are essential tools in calculus, allowing us to find the rate at which a function is changing at any given point. There are multiple methods to differentiate various types of functions, and choosing the right technique is crucial for simplifying the process.

- **Logarithmic Differentiation**: Particularly useful for functions where the independent variable appears in multiple factors. By taking the natural logarithm of both sides, complex products or quotients can be transformed into sums and differences, making them easier to differentiate. - **Chain Rule**: Used when differentiating composite functions. When you have a function inside another function, you differentiate the outer function and multiply it by the derivative of the inner function.

Using these techniques effectively means being able to recognize when to apply each method, based on the structure of the function you are working with.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\approx 2.71828\). It is a fundamental function in calculus because it naturally arises in the processes involving continuous growth or decay.

- **Properties of Natural Logarithms**: - \(\ln(ab) = \ln a + \ln b\): This property allows the simplification of the logarithm of a product into the sum of logarithms, which can hugely simplify differentiation. - \(\ln(a^b) = b\ln a\): This property is particularly useful when dealing with powers.- **Differentiating \(\ln x\)**: The derivative of \(\ln x\) with respect to \(x\) is simply \(\frac{1}{x}\). This property is integral to many differentiation problems, particularly when using logarithmic differentiation.

The natural logarithm thus plays a crucial role in simplifying functions to facilitate easier differentiation.

Product Rule

The product rule is a fundamental technique used to differentiate products of two or more functions. It is expressed mathematically as follows: \[\frac{d}{dt}(u(t)v(t)) = u'(t)v(t) + u(t)v'(t)\]

In the case of more than two functions, you can extend the product rule similarly. Each function is differentiated in turn while the other functions remain constant. This produces a sum of terms where each term has one of the functions differentiated and the others not.

• **Two Functions**: If \(u(t)\) and \(v(t)\) are your functions, differentiate each in turn, keeping the other unchanged for each part of the sum.
• **More than Two Functions**: When dealing with three or more functions, extend the rule by considering each function to be differentiated once, pairing with the product of the derivatives of the others as constant.
  
  
  Practicing the product rule with simple functions quickly builds the intuition needed for more complex problems.

Derivative Computation

The computation of derivatives is a key task in calculus, which involves finding the derivative or rate of change of a function. It can involve straightforward applications of differentiation rules like the product or chain rule, or more advanced techniques like logarithmic differentiation.

- **Basic Steps in Derivative Computation**: - Recognize the form of function: Determine if the function is a basic polynomial, exponential, logarithmic, or a combination of these, to choose the right approach. - Apply the appropriate differentiation rules: Use the product rule, quotient rule, or chain rule as necessary based on the function's composition. - Simplify the result: Often, it requires algebraic manipulation to express the derivative in its simplest form.

By following these steps methodically, students can master the art of computing derivatives, paving the way for understanding more complex calculus concepts.