deritive of: f (x) = (x2 + 1)(x − 1) x + 4

To find the derivative of the function:

\[
f(x) = \frac{(x^2 + 1)(x - 1)}{x + 4}
\]

we can use the **quotient rule** for derivatives, which states:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

where \( u = (x^2 + 1)(x - 1) \) and \( v = x + 4 \). We need to find the derivatives \( u' \) and \( v' \) to proceed.

### Step 1: Differentiate \( u = (x^2 + 1)(x - 1) \)
Use the **product rule** on \( u \):

1. Let \( g(x) = x^2 + 1 \) and \( h(x) = x - 1 \).
2. Then \( u = g(x) \cdot h(x) \).
3. By the product rule, \( u' = g' \cdot h + g \cdot h' \).

#### Compute \( g' \) and \( h' \):
- \( g'(x) = 2x \)
- \( h'(x) = 1 \)

#### Substitute back to find \( u' \):
\[
u' = (2x)(x - 1) + (x^2 + 1)(1) = 2x(x - 1) + (x^2 + 1)
\]
\[
= 2x^2 - 2x + x^2 + 1 = 3x^2 - 2x + 1
\]

### Step 2: Differentiate \( v = x + 4 \)
\[
v' = 1
\]

### Step 3: Apply the Quotient Rule
Now, substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula:

\[
f'(x) = \frac{(x + 4)(3x^2 - 2x + 1) - (x^2 + 1)(x - 1)}{(x + 4)^2}
\]

To fully simplify, expand the terms in the numerator:

1. Expand \((x + 4)(3x^2 - 2x + 1)\).
2. Expand \((x^2 + 1)(x - 1)\).
3. Combine like terms.

Would you like me to continue with these expansions, or is this enough detail?

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### Follow-up Questions

1. How does the product rule apply to derivatives in general?
2. What other functions could be differentiated using the quotient rule?
3. How does the derivative help describe the behavior of the function?
4. How would this derivative be applied in an optimization problem?
5. Could we use a different method to solve this derivative?

### Tip
When working with complex functions, break them down into smaller components to simplify the derivative process.

derivative of: f(x) = (x^2 + 1)(x − 1) / (x + 4)

This page provides a detailed solution for finding the derivative of f(x) = (x^2 + 1)(x - 1) / (x + 4) using the quotient rule and product rule in calculus. Explore step-by-step differentiation, formulas, and key concepts in derivatives suitable for Grades 11-12 students.

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