Problem 110 - Calculus and its applications

Question

Differentiate. $$ f(t)=\ln \left[\left(t^{3}+3\right)\left(t^{2}-1\right)\right] $$

Step-by-Step Solution

Answer

The derivative is \(f'(t) = \frac{3t^2}{t^3 + 3} + \frac{2t}{t^2 - 1}\).

Step 1: Apply the Logarithm Property

The first step is to simplify the given function using the property of logarithms: \[\ln(a \cdot b) = \ln(a) + \ln(b).\] Apply this property to the function:\[f(t) = \ln[(t^3 + 3)(t^2 - 1)] = \ln(t^3 + 3) + \ln(t^2 - 1).\]

Step 2: Differentiate Each Term Separately

Now differentiate each term separately with respect to t. We'll use the derivative of \(\ln(u)\), which is \(\frac{1}{u} \cdot \frac{du}{dt}\). Let's find the derivatives:- For \(\ln(t^3 + 3)\): \[ \frac{d}{dt}[\ln(t^3 + 3)] = \frac{1}{t^3 + 3} \cdot 3t^2. \]- For \(\ln(t^2 - 1)\): \[ \frac{d}{dt}[\ln(t^2 - 1)] = \frac{1}{t^2 - 1} \cdot 2t. \]

Step 3: Combine the Derivative Results

Now, combine the derivatives found in Step 2 to get the derivative of the entire function:\[f'(t) = \frac{3t^2}{t^3 + 3} + \frac{2t}{t^2 - 1}.\]

Key Concepts

Logarithm Properties

Logarithm properties are incredibly handy when simplifying complex expressions, especially before taking derivatives. A key logarithm property is that the logarithm of a product can be expressed as the sum of the logarithms of its factors. This is written as: \( \ln(a \cdot b) = \ln(a) + \ln(b) \). In the context of differentiation, this property allows us to break down complicated logarithmic expressions into simpler, more manageable pieces.
For example, with our function \( f(t) = \ln\left[(t^3 + 3)(t^2 - 1)\right] \), applying the product-to-sum property of logarithms simplifies it to \( \ln(t^3 + 3) + \ln(t^2 - 1) \).
A handy mnemonic to remember this property is “log of a product, sum it up!” This simplification step is crucial because it makes further differentiation steps much easier by splitting the task into smaller parts.

Chain Rule

The chain rule is a fundamental tool in calculus for taking the derivative of composite functions. When you have a function within another function, like \( \ln(u) \), where \( u \) is itself a function of \( t \), the chain rule tells us how to differentiate it.
The chain rule formula is: \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). In simpler terms, you differentiate the outer function as if the inner function were just a simple variable, then multiply by the derivative of the inner function.
In our example, for differentiating \( \ln(t^3 + 3) \), we apply the chain rule by taking the derivative of \( \ln(u) \) which is \( \frac{1}{u} \), and multiplying it by the derivative of the inner function \( t^3 + 3 \), which results in \( 3t^2 \). The chain rule makes handling nested functions much more systematic.

Product Rule

The product rule helps when differentiating a product of two functions. Though in our specific exercise we used logarithm properties to avoid the direct use of the product rule, it's still an essential concept.
The product rule states that if you have a function \( h(t) = u(t) \cdot v(t) \), the derivative is: \( h'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t) \). This rule allows us to properly account for changes in both functions involved in the product, producing an accurate derivative.
Understanding the product rule is particularly beneficial when differentiating products directly, without the simplification of logarithm properties, especially when the product consists of functions that cannot be easily combined or transformed into a sum.