Question

Differentiate the following functions: \(\frac{x^{3}+a^{3}}{x+a}\)

Short answer

Question: Find the derivative of the function \(y = \frac{x^{3}+a^{3}}{x+a}\). Answer: The derivative of the function is \(y' = \frac{2x^{3} + 3ax^{2} - a^{3}}{(x+a)^{2}}\).

Step-by-step solution

Identify the functions that make up the quotient

In this case, the given function is \(\frac{x^{3}+a^{3}}{x+a}\). Therefore, we have: Function \(f(x) = x^{3}+a^{3}\) and function \(g(x) = x+a\)

Differentiate both functions separately

Differentiate \(f(x)\) and \(g(x)\) with respect to x: \(f'(x) = \frac{d}{dx}(x^{3}+a^{3}) = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(a^{3}) = 3x^{2} + 0 = 3x^{2}\) \(g'(x) = \frac{d}{dx}(x+a) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1\)

Apply the quotient rule for differentiation

The quotient rule states that if we have a function which is a quotient of two other functions, say \(u(x)=\frac{f(x)}{g(x)}\), then the derivative of \(u(x)\) is given by \ \(u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^{2}}\) Applying the quotient rule to the given function: \(u'(x) = \frac{(3x^{2})(x+a) - (x^{3}+a^{3})(1)}{(x+a)^{2}}\)

Simplify the expression for the derivative

Simplify the numerator of the expression: \(u'(x) = \frac{3x^{3} + 3ax^{2} - x^{3} - a^{3}}{(x+a)^{2}}\) Combine the terms in the numerator: \(u'(x) = \frac{2x^{3} + 3ax^{2} - a^{3}}{(x+a)^{2}}\) This is the derivative of the given function.

Key concepts

Quotient Rule

In differential calculus, one common technique for differentiating quotients is the Quotient Rule. This rule helps us find the derivative of a function that is the ratio between two other functions. The quotient rule can be remembered through the following formula:
\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\]This means to differentiate a quotient \(\frac{f(x)}{g(x)}\), we:

• Differentiate the top function \(f(x)\) to get \(f'(x)\).
• Differentiate the bottom function \(g(x)\) to get \(g'(x)\).
• Plug these results into the formula mentioned above.
• Simplify the resulting expression.

It's important to remember that the denominator \(g(x)^2\) in the quotient rule formula ensures that the division by zero doesn't occur, assuming \(g(x) eq 0\). This method is particularly useful when we are given complex fractional functions, like the one we have here: \(\frac{x^3 + a^3}{x + a}\).

Function Differentiation

Function differentiation involves finding the rate at which a function's value changes with respect to changes in its input variable. In our exercise, we are differentiating two separate components of the given function. Here's a simple breakdown:

• For the function \(f(x) = x^3 + a^3\), the differentiation process involves calculating the derivative of each term. Using the power rule, we know \(\frac{d}{dx}(x^3) = 3x^2\) and the derivative of a constant, \(a^3\), is \(0\).
• The function \(g(x) = x + a\) is simpler. The derivative of \(x\) is \(1\), and since \(a\) is a constant, \(\frac{d}{dx}(a) = 0\). Thus, \(g'(x) = 1\).

Differentiation uncovers the instantaneous rate of change or the slope of the function at any given point, which is fundamental in understanding the behavior of mathematical models in real-world applications.

Mathematics Education

In mathematics education, understanding core concepts like the Quotient Rule and function differentiation is crucial for students. These concepts form the basis for more advanced studies in calculus and other mathematical disciplines. Educators focus on ensuring comprehension through:

• Interactive learning, which involves solving problems in groups or using technology-based applications.
• Head-on methodology, breaking down tough equations into approachable steps. For example, working through each component as we did with \(f(x) = x^3 + a^3\) and \(g(x) = x + a\).
• Using visual aids, which can help students visualize changes and understand concepts like slope and rate of change.
• Practicing with real-life examples that show how calculus is applied in fields like physics, engineering, and economics.

By building a solid foundation in these key areas, students are better prepared for tackling complex mathematical challenges both in academics and in various professional fields.