Questions: Differentiate (x^3 + 2x) / (x - 1)

Solution Steps

To differentiate the given function, we will use the quotient rule. The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then its derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

In this case, \( g(x) = x^3 + 2x \) and \( h(x) = x - 1 \). We will find the derivatives \( g'(x) \) and \( h'(x) \), and then apply the quotient rule.

We are given the function \( f(x) = \frac{x^3 + 2x}{x - 1} \). To differentiate this function, we identify:

• \( g(x) = x^3 + 2x \)
• \( h(x) = x - 1 \)

The derivatives are:

• \( g'(x) = 3x^2 + 2 \)
• \( h'(x) = 1 \)

The quotient rule for differentiation is given by:

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

Substituting the derivatives and functions, we have:

\[ f'(x) = \frac{(3x^2 + 2)(x - 1) - (x^3 + 2x)(1)}{(x - 1)^2} \]

Simplifying the expression, we get:

\[ f'(x) = \frac{(3x^3 - 3x^2 + 2x - 2) - (x^3 + 2x)}{(x - 1)^2} \]

This simplifies further to:

\[ f'(x) = \frac{2x^3 - 3x^2 - 2}{x^2 - 2x + 1} \]

Final Answer