Question

(a) Use the Quotient Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Quotient Rule. (c) Show that the answers from \((\mathrm{a})\) and \((\mathrm{b})\) are equivalent. $$f(x)=\frac{x^{2}+3}{x}$$

Short answer

Both methods give the derivative: \( f'(x) = 1 - \frac{3}{x^2} \).

Step-by-step solution

Differentiate Using the Quotient Rule

The quotient rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). For \( f(x) = \frac{x^2 + 3}{x} \), we have \( u(x) = x^2 + 3 \) and \( v(x) = x \). First, find \( u'(x) = 2x \) and \( v'(x) = 1 \). Then, apply the quotient rule:

Applying the Quotient Rule

Substitute into the quotient rule formula: \[ f'(x) = \frac{(2x)(x) - (x^2 + 3)(1)}{x^2} = \frac{2x^2 - x^2 - 3}{x^2} = \frac{x^2 - 3}{x^2}. \]

Simplify the Function Algebraically

Rewrite the given function \( f(x) = \frac{x^2 + 3}{x} \) by separating the terms in the numerator: \[ f(x) = \frac{x^2}{x} + \frac{3}{x} = x + 3x^{-1}. \]This algebraic manipulation simplifies the expression into a form easier to differentiate.

Differentiate the Simplified Function

Differentiate \( f(x) = x + 3x^{-1} \). The derivative of \( x \) is \( 1 \) and the derivative of \( 3x^{-1} \) is \( -3x^{-2} \). Thus, \[ f'(x) = 1 - 3x^{-2}. \]

Verify Equivalence of Answers

We obtained two answers: 1. From quotient rule: \( f'(x) = \frac{x^2 - 3}{x^2} \) 2. From algebraic manipulation: \( f'(x) = 1 - 3x^{-2} \).Simplify the first answer: \[ \frac{x^2 - 3}{x^2} = \frac{x^2}{x^2} - \frac{3}{x^2} = 1 - \frac{3}{x^2}. \]Both derivatives are expressed as \( 1 - 3x^{-2} \), confirming they are equivalent.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. In simple terms, it's about understanding how one quantity changes in relation to another. When we differentiate a function, we obtain its derivative, which provides this rate of change.

When applying differentiation to a function like \( f(x) = \frac{x^2 + 3}{x} \), we often use rules that simplify the process. One powerful tool is the Quotient Rule, specifically designed for functions expressed as fractions. The Quotient Rule formula is given by:

• \( u(x) \): the numerator function.
• \( v(x) \): the denominator function.
• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

By differentiating the numerator and denominator separately and applying the formula, we determine how \( f(x) \) changes over its domain without directly manipulating the original expression.

Algebraic Manipulation

Algebraic manipulation involves rewriting expressions in more useful or simplified forms. This process can make differentiating a function easier, especially when the function is a complex fraction.

In the case of the function \( f(x) = \frac{x^2 + 3}{x} \), we can simplify it by splitting the numerator. This gives us:

• \( \frac{x^2}{x} = x \)
• \( \frac{3}{x} = 3x^{-1} \)

Thus, the function becomes \( f(x) = x + 3x^{-1} \). This transformation not only tidies up the expression but also makes revealing the derivative straightforward. Simplified expressions enable faster computations and help avoid errors typical of less transparent forms in differentiation.

Derivative Verification

Checking the correctness of derivatives is crucial and is often achieved through derivative verification. By comparing derivatives obtained through different methods, we ensure accuracy and consistency.

In our example function \( f(x) = \frac{x^2 + 3}{x} \), we derived the following:

• Using the Quotient Rule: \( f'(x) = \frac{x^2 - 3}{x^2} \)
• Using algebraic manipulation: \( f'(x) = 1 - 3x^{-2} \)

To verify, simplify the result from the Quotient Rule:

• \( \frac{x^2 - 3}{x^2} = \frac{x^2}{x^2} - \frac{3}{x^2} = 1 - \frac{3}{x^2} \)

Both methods yield \( f'(x) = 1 - 3x^{-2} \), confirming their equivalence. Verification provides confidence in our solution and illustrates the reliability of using different approaches to differentiation.