Problem 11 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=\frac{3 x+4}{x^{2}+1}\)

Step-by-Step Solution

Answer

\(f^{\prime}(x) = \frac{-3x^2 - 8x + 3}{(x^2 + 1)^2}\)

Step 1: Recall the Quotient Rule for Derivatives

The function \(f(x)\) is a quotient of two functions, \(u(x) = 3x + 4\) and \(v(x) = x^2 + 1\). The derivative of a quotient \(\frac{u}{v}\) is given by the quotient rule: \[f^{\prime}(x) = \frac{u^{\prime}(x) v(x) - u(x) v^{\prime}(x)}{(v(x))^2}\]where \(u^{\prime}(x)\) and \(v^{\prime}(x)\) are the derivatives of \(u(x)\) and \(v(x)\) respectively.

Step 2: Differentiating the Numerator and Denominator

Calculate \(u^{\prime}(x)\) and \(v^{\prime}(x)\).- For \(u(x) = 3x + 4\), the derivative is: \[u^{\prime}(x) = 3\]- For \(v(x) = x^2 + 1\), the derivative is: \[v^{\prime}(x) = 2x\]

Step 3: Apply the Quotient Rule

Use the quotient rule to differentiate \(f(x)\): \[f^{\prime}(x) = \frac{3(x^2 + 1) - (3x + 4)(2x)}{(x^2 + 1)^2}\]Simplify the expression in the numerator to find the actual derivative.

Step 4: Simplify the Expression

Compute and simplify the expression in the numerator:- Expand both parts: \[3(x^2 + 1) = 3x^2 + 3\] \[(3x + 4)(2x) = 6x^2 + 8x\]- Subtract the results: \[3x^2 + 3 - (6x^2 + 8x) = 3x^2 + 3 - 6x^2 - 8x\] \[= -3x^2 - 8x + 3\]Thus, \[f^{\prime}(x) = \frac{-3x^2 - 8x + 3}{(x^2 + 1)^2}\]

Step 5: Conclusion

The derivative of the function \(f(x) = \frac{3x + 4}{x^2 + 1}\) is given by the simplified expression\[f^{\prime}(x) = \frac{-3x^2 - 8x + 3}{(x^2 + 1)^2}\]

Key Concepts

Derivatives

In calculus, derivatives help us understand how a function behaves by telling us its rate of change. Imagine you're driving a car. The speedometer shows how fast you're going at any given moment. Similarly, a derivative gives us the function's rate of change at a particular point.
Derivatives are especially useful in

• understanding motion, such as speeds and acceleration,
• finding maximum and minimum values of a function,
• and solving real-world optimization problems.

To find the derivative, we typically use rules developed from limits. The task becomes easier with the aid of differentiation techniques, like the power rule, product rule, and, as we'll see in this example, the quotient rule.

Rational Functions

Rational functions are fractions made up of two polynomials. In this exercise, the function is \[f(x) = \frac{3x + 4}{x^2 + 1}\]This means it consists of

• a numerator, \(3x + 4\),
• and a denominator, \(x^2 + 1\).

Rational functions can be a bit tricky to deal with, especially when finding derivatives, because of their fraction format. Still, they are rich in real-world applications, appearing in fields ranging from physics to economics.
The process of finding the derivative of such a function typically involves applying the quotient rule, which is a useful tool when working with functions like these.

Differentiation Techniques

Differentiation techniques provide a toolbox for finding derivatives. When tasked with finding the derivative of more complex functions, such as rational ones, these techniques become vital. In our example, we've used the quotient rule, which is specifically designed for dividing two functions.
The quotient rule states that if you have a function \(f(x) = \frac{u(x)}{v(x)}\), the derivative \(f'(x)\) is given by:\[f^{\prime}(x) = \frac{u^{\prime}(x) v(x) - u(x) v^{\prime}(x)}{(v(x))^2}\]In this approach:

• \(u(x)\) and \(v(x)\) are the numerator and denominator respectively,
• and \(u^{\prime}(x)\) and \(v^{\prime}(x)\) are their derivatives.

The quotient rule helps break down the process into simpler parts:
1. Find the derivatives of the numerator and denominator separately.
2. Substitute into the quotient rule formula.
3. Simplify the resulting expression.
Employing these techniques ensures accuracy in calculus, allowing us to solve problems efficiently and with precision.