Problem 45 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(t)=\frac{t^{2}+2 t-1}{t^{2}+t-3} $$

Step-by-Step Solution

Answer

The derivative is \( f'(t) = \frac{-t^2 - 6t - 5}{(t^2 + t - 3)^2} \).

Step 1: Identify the Functions in the Quotient

The given function is a quotient \( f(t) = \frac{u(t)}{v(t)} \) where \( u(t) = t^2 + 2t - 1 \) and \( v(t) = t^2 + t - 3 \).

Step 2: Write the Quotient Rule Formula

The Quotient Rule states that the derivative of a quotient \( \frac{u(t)}{v(t)} \) is \( \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2} \).

Step 3: Find the Derivative of the Numerator

The numerator is \( u(t) = t^2 + 2t - 1 \). Differentiating term by term: \( u'(t) = 2t + 2 \).

Step 4: Find the Derivative of the Denominator

The denominator is \( v(t) = t^2 + t - 3 \). Differentiating term by term: \( v'(t) = 2t + 1 \).

Step 5: Apply the Quotient Rule

Substitute the derivatives found into the quotient rule formula: \[f'(t) = \frac{(2t + 2)(t^2 + t - 3) - (t^2 + 2t - 1)(2t + 1)}{(t^2 + t - 3)^2}.\]

Step 6: Expand the Numerator

Calculate \((2t + 2)(t^2 + t - 3) = 2t^3 + 2t^2 - 6t + 2t^2 + 2t - 6\) which simplifies to \(2t^3 + 4t^2 - 4t - 6\).

Step 7: Expand the Second Term of Numerator

Calculate \((t^2 + 2t - 1)(2t + 1) = 2t^3 + t^2 + 4t^2 + 2t - 2t - 1\) which simplifies to \(2t^3 + 5t^2 + 2t - 1\).

Step 8: Simplify the Numerator

Combine the expanded terms: \(2t^3 + 4t^2 - 4t - 6 - (2t^3 + 5t^2 + 2t - 1)\), which simplifies to \(-t^2 - 6t - 5\).

Step 9: Write the Simplified Derivative

The simplified derivative using the quotient rule is \[f'(t) = \frac{-t^2 - 6t - 5}{(t^2 + t - 3)^2}.\]

Key Concepts

Derivative Calculation

The derivative of a function measures how the function's value changes as its input changes. Calculating derivatives is an essential skill in calculus and helps analyze the behaviour of functions. In our exercise, the derivative of a function of the form \( f(t) = \frac{u(t)}{v(t)} \) can be found using the Quotient Rule.
The derivative calculation involves a formula where we take the derivatives of both the numerator and the denominator separately. This process requires attention to detail, as each function might involve different terms or powers. When finding these derivatives, break down the expression term by term.

• For example, with \( u(t) = t^2 + 2t - 1 \), the derivative \( u'(t) = 2t + 2 \) is found by differentiating each term.
• Similarly, for \( v(t) = t^2 + t - 3 \), \( v'(t) = 2t + 1 \).

With these derivatives, we use the Quotient Rule to calculate the overall derivative of the function.

Simplifying Algebraic Expressions

Simplification is a critical step in making derivative expressions easier to understand and use. Once the derivatives are calculated using the Quotient Rule, the numerator often undergoes expansion and combination of like terms.
This involves multiplying each term in one polynomial by each term in another, which can initially lead to a complex mix of terms. Let's break down the simplification process:

• Start by expanding expressions inside parentheses, ensuring that each term multiplies correctly.
• Carefully add or subtract these expanded forms, paying close attention to negative signs.

In our solution, after expanding and combining terms, the numerator simplifies from a complicated expression to \(-t^2 - 6t - 5\). This result makes it feasible to further analyze or apply the derivative as needed.

Differentiation Techniques

Differentiation is a set of techniques used to find the derivative of a function. One such technique is the Quotient Rule, used when handling a function that is a quotient of two other functions, \( \frac{u(t)}{v(t)} \). This rule is powerful because it directly applies to any quotient, allowing you to find derivatives systematically.
In this exercise:

• The application of the Quotient Rule is essential; its formula is \( \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2} \), ensuring the result considers both parts of the quotient.
• The process involves initially understanding and breaking down the functions involved, differentiating them, and then substituting these values back into the Quotient Rule.

Each differentiation technique, including this one, provides a framework for systematically approaching problems, ensuring derivatives are found accurately and efficiently. Having these techniques in your math toolkit allows you not only to solve textbook problems but also to tackle real-world scenarios involving rates of change.