Problem 63 - Thomas Calculus

Question

In Exercises \(57-70\) , 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

The derivative is \( 3t^2 + 6t + 2 \).

Step 1: Apply the Natural Logarithm

Apply the natural logarithm to both sides of the equation: \[ \ln y = \ln(t(t+1)(t+2)) \] Using the property of logarithms, this becomes: \[ \ln y = \ln t + \ln(t+1) + \ln(t+2) \]

Step 2: Differentiate Implicitly with Respect to t

Differentiate both sides with respect to the independent variable \( t \) using implicit differentiation: \[ \frac{d}{dt}(\ln y) = \frac{d}{dt}(\ln t + \ln(t+1) + \ln(t+2)) \] This results in: \[ \frac{1}{y} \frac{dy}{dt} = \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \]

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

Solve for \( \frac{dy}{dt} \) by multiplying through by \( y \): \[ \frac{dy}{dt} = y \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right) \] Since \( y = t(t+1)(t+2) \), substitute back: \[ \frac{dy}{dt} = t(t+1)(t+2) \left( \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} \right) \]

Step 4: Simplify the Expression

Distribute and simplify the expression: \[ \frac{dy}{dt} = (t+1)(t+2) + t(t+2) + t(t+1) \] This simplifies to: \[ \frac{dy}{dt} = t^2 + 3t + 2 + t^2 + 2t + t^2 + t = 3t^2 + 6t + 2 \]

Key Concepts

Implicit Differentiation

Implicit differentiation allows us to differentiate equations where the variables are intertwined, meaning not solved explicitly for one variable. To perform implicit differentiation, both sides of the equation are differentiated, treating one variable as a function of the other. In our case, when we differentiated \( \ln y = \ln t + \ln(t+1) + \ln(t+2) \) with respect to \( t \), we treated \( y \) as a function of \( t \).
This is why we introduced \( \frac{dy}{dt} \) on the left side, as it represents how \( y \) changes in response to slight changes in \( t \). Implicit differentiation is especially useful when dealing with circular and exponential relationships, making it a powerful tool in calculus.

Natural Logarithm

Natural logarithm, denoted by \( \ln \), is a logarithm to the base of the Euler's number, \( e \). It's a fundamental function in calculus that simplifies the process of differentiation by turning multiplicative relationships into additive ones.
In this exercise, applying the natural logarithm to both sides means we utilize that \( \ln(abc) = \ln a + \ln b + \ln c \). This transformation simplifies handling products of functions, enabling easier derivative calculations.

• With \( \ln y = \ln(t(t+1)(t+2)) \), we simplify further to \( \ln y = \ln t + \ln(t+1) + \ln(t+2) \).


• This addition breaks down the function, paving the way for straightforward differentiation using rules for logs.

Derivative Calculation

Derivative calculation involves finding the rate at which a function changes. For logarithmic differentiation, it is a method where we apply the natural logarithm to simplify the differentiation of complex functions. In the exercise, once the equation was log-transformed, different pieces of the function were differentiated separately.

• For \( \frac{d}{dt}(\ln y) \), it turned into \( \frac{1}{y} \frac{dy}{dt} \)


• For \( \frac{d}{dt}(\ln t) \), it resulted in \( \frac{1}{t} \).
• Adding the outputs together, derived the expression for \( \frac{dy}{dt} \).

Derivative calculation is crucial for uncovering the relationship between variables, especially in physical systems and dynamic processes.

Product Rule

The product rule is a differential calculus principle used to find the derivative of a product of two or more functions. In straightforward terms, if you have a function \( f(x) \times g(x) \), the product rule tells us how to derive it: \( f'(x)g(x) + f(x)g'(x) \).
However, in logarithmic differentiation, we bypass the complexity of the product rule directly on the original function by transforming the product into a sum of logarithms. In our example, differentiating \( \ln(t) + \ln(t+1) + \ln(t+2) \) effectively used the simplification afforded by the natural logarithm. This helps avoid complicated multi-step product rules, which can be tricky with compounded terms.

• Once differentiation was applied, we essentially circled back to simplifying expressions like \( (t+1)(t+2) + t(t+2) + t(t+1) \).


• Comparing results, these simplify similar handling as direct product rule would, but with fewer steps involved.

This strategic use of logarithmic differentiation streamlines solving otherwise challenging derivative problems involving products.