Problem 20 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions. $$f(t)=\frac{t^{2}-1}{t^{2}+t-2}$$

Step-by-Step Solution

Answer

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

Step 1: Write Down the Quotient Rule

To find the derivative of a function of the form \( f(t) = \frac{g(t)}{h(t)} \), we use the quotient rule. The formula is \( (\frac{g(t)}{h(t)})' = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2} \). Here, \( g(t) = t^2 - 1 \) and \( h(t) = t^2 + t - 2 \).

Step 2: Differentiate the Numerator

Find the derivative of the numerator \( g(t) = t^2 - 1 \). The derivative is \( g'(t) = 2t \).

Step 3: Differentiate the Denominator

Find the derivative of the denominator \( h(t) = t^2 + t - 2 \). The derivative is \( h'(t) = 2t + 1 \).

Step 4: Apply the Quotient Rule Formula

Now plug the derivatives and the original functions into the quotient rule: \[ (\frac{t^2-1}{t^2+t-2})' = \frac{(2t)(t^2+t-2) - (t^2-1)(2t+1)}{(t^2+t-2)^2} \].

Step 5: Simplify the Numerator

Expand and simplify the terms in the numerator: \( (2t)(t^2+t-2) = 2t^3 + 2t^2 - 4t \) and \( (t^2-1)(2t+1) = 2t^3 + t^2 - 2t - 1 \). Subtract these to get: \( (2t^3 + 2t^2 - 4t) - (2t^3 + t^2 - 2t - 1) = t^2 - 2t + 1 \).

Step 6: Write the Final Simplified Derivative

Substituting back into the formula, the derivative is: \[ f'(t) = \frac{t^2 - 2t + 1}{(t^2 + t - 2)^2} \]. This is the simplified form of the derivative.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It is essentially a measure of the rate at which a quantity changes. For a function \( f(t) \), the derivative, denoted as \( f'(t) \), captures the slope of the tangent line at any given point on the graph of \( f(t) \). In this particular exercise, where the function is given as a fraction, we apply a derivative technique called the Quotient Rule.
Understanding derivatives is key to solving many problems in calculus, as they tell us about the function's behavior, such as where it increases or decreases, and can inform us about the shape of the graph of the function. Calculating the derivative helps us understand the underlying rate of change in relation to its variables.
When dealing with complex functions like fractions, using specific rules like the Quotient Rule helps you break down the function into more straightforward parts, making it easier to take derivatives.

Numerator and Denominator Differentiation

When applying the Quotient Rule, it's crucial first to differentiate both the numerator and the denominator individually. For our given function \( f(t) = \frac{t^2 - 1}{t^2 + t - 2} \), we need to find \( g'(t) \) and \( h'(t) \) where \( g(t) = t^2 - 1 \) and \( h(t) = t^2 + t - 2 \).
Differentiation of these parts is straightforward as they are polynomials:

• The derivative of the numerator, \( g(t) = t^2 - 1 \), is \( 2t \).
• The derivative of the denominator, \( h(t) = t^2 + t - 2 \), is \( 2t + 1 \).

Each part is simplified independently before being plugged back into the Quotient Rule formula, which combines them to give the overall derivative of the function.

Simplification of Derivatives

After applying the Quotient Rule and obtaining the derivatives of the numerator and denominator, the next essential step is simplifying the result. Simplification is a critical part of finding derivatives, especially for making them more manageable and easier to interpret.
In our example, the formula derived from the Quotient Rule is:\[ f'(t) = \frac{(2t)(t^2+t-2) - (t^2-1)(2t+1)}{(t^2+t-2)^2} \]The task here is to simplify the expression in the numerator. First, we expand each term:

• \( (2t)(t^2+t-2) = 2t^3 + 2t^2 - 4t \)
• \( (t^2-1)(2t+1) = 2t^3 + t^2 - 2t - 1 \)

Subtracting these products gives us \( t^2 - 2t + 1 \). We then substitute this back into the formula for \( f'(t) \), resulting in the simplified form:\[ f'(t) = \frac{t^2 - 2t + 1}{(t^2 + t - 2)^2} \]
This form is often preferred because it is easier to evaluate and analyze for various inputs of \( t \).