Find f prime left parenthesis x right parenthesis and f double prime left parenthesis x right parenthesis. 
​f(x)equalsStartFraction x squared minus 3 x Over x plus 1 EndFraction
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Part 1
f prime left parenthesis x right parenthesisequals
  
StartFraction left parenthesis x plus 3 right parenthesis left parenthesis x minus 1 right parenthesis Over left parenthesis x plus 1 right parenthesis squared EndFraction
Part 2
f double prime left parenthesis x right parenthesisequals
  
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Let's go through the steps to find the first and second derivatives of the function \( f(x) = \frac{x^2 - 3x}{x + 1} \).

### Step 1: Finding \( f'(x) \)
We use the **quotient rule** for derivatives. The quotient rule is given by:

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

For the function \( f(x) = \frac{x^2 - 3x}{x + 1} \), we define:
- \( u(x) = x^2 - 3x \)
- \( v(x) = x + 1 \)

Now, we find \( u'(x) \) and \( v'(x) \):
- \( u'(x) = 2x - 3 \)
- \( v'(x) = 1 \)

Using the quotient rule:

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

Simplifying the numerator:

\[
(2x - 3)(x + 1) = 2x^2 + 2x - 3x - 3 = 2x^2 - x - 3
\]
\[
(x^2 - 3x)(1) = x^2 - 3x
\]
So the numerator becomes:

\[
(2x^2 - x - 3) - (x^2 - 3x) = 2x^2 - x - 3 - x^2 + 3x = x^2 + 2x - 3
\]

Thus, the first derivative is:

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

We can factor the numerator:

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

This matches the given result for \( f'(x) \).

### Step 2: Finding \( f''(x) \)
Now we need to find the second derivative, \( f''(x) \). To do this, we will differentiate \( f'(x) = \frac{(x + 3)(x - 1)}{(x + 1)^2} \) using the quotient rule again.

Let:
- \( u(x) = (x + 3)(x - 1) \)
- \( v(x) = (x + 1)^2 \)

Now, we find \( u'(x) \) and \( v'(x) \):
- \( u(x) = x^2 + 2x - 3 \), so \( u'(x) = 2x + 2 \)
- \( v(x) = (x + 1)^2 \), so \( v'(x) = 2(x + 1) \)

Now apply the quotient rule:

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

Simplifying the numerator:

\[
(2x + 2)(x + 1)^2 = (2x + 2)(x^2 + 2x + 1) = 2x^3 + 4x^2 + 2x + 2x^2 + 4x + 2 = 2x^3 + 6x^2 + 6x + 2
\]
\[
(x^2 + 2x - 3)(2(x + 1)) = 2(x^2 + 2x - 3)(x + 1) = 2(x^3 + 3x^2 - x - 3) = 2x^3 + 6x^2 - 2x - 6
\]
Now subtract the two expressions:

\[
(2x^3 + 6x^2 + 6x + 2) - (2x^3 + 6x^2 - 2x - 6) = 8x + 8
\]

Thus, the second derivative is:

\[
f''(x) = \frac{8(x + 1)}{(x + 1)^4} = \frac{8}{(x + 1)^3}
\]

### Final Result:

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

Let me know if you would like more details or have questions about any step.

---

### Relative Questions:
1. How do you apply the quotient rule for finding the derivative of a function?
2. Can you explain how to simplify after applying the quotient rule?
3. What is the significance of the second derivative \( f''(x) \)?
4. How do you handle the product rule when differentiating?
5. What is the interpretation of critical points found from \( f'(x) = 0 \)?

### Tip:
When simplifying derivatives, factoring terms can make the result clearer and help check for mistakes.

Find f prime left parenthesis x right parenthesis and f double prime left parenthesis x right parenthesis. f(x) equals StartFraction x squared minus 3 x Over x plus 1 EndFraction

Learn how to find the first and second derivatives of the function f(x) = (x^2 - 3x) / (x + 1) using the quotient rule. This detailed solution walks through the differentiation steps and simplifies the results for both f'(x) and f''(x). Suitable for college-level calculus students.

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