We need to differentiate the function:

$f(x) = \frac{(x - 1)^3}{(x^2 + x)^5}$

Step 1: Use the Quotient Rule

The quotient rule states that if $f(x) = \frac{g(x)}{h(x)}$, then:

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

where:

• $g(x) = (x - 1)^3$
• $h(x) = (x^2 + x)^5$

Step 2: Compute $g'(x)$

Using the power rule:

$g(x) = (x - 1)^3$

Differentiate using the chain rule:

$g'(x) = 3(x - 1)^2$

Step 3: Compute $h'(x)$

First, rewrite:

$h(x) = (x^2 + x)^5$

Using the chain rule:

$h'(x) = 5(x^2 + x)^4 \cdot (2x + 1)$

Step 4: Apply the Quotient Rule

$f'(x) = \frac{3(x - 1)^2 (x^2 + x)^5 - (x - 1)^3 \cdot 5(x^2 + x)^4 (2x + 1)}{(x^2 + x)^{10}}$

Factor out the common terms:

$f'(x) = \frac{(x - 1)^2 (x^2 + x)^4 \left[ 3(x^2 + x) - 5(x - 1)(2x + 1) \right]}{(x^2 + x)^{10}}$

Simplify the bracketed term:

$3(x^2 + x) - 5(x - 1)(2x + 1)$

Expand $5(x - 1)(2x + 1)$:

= 5(2x^2 + x - 2x - 1) = 5(2x^2 - 1) = 10x^2 - 5$$ So, $$3(x^2 + x) - (10x^2 - 5)$$ Expanding: $$3x^2 + 3x - 10x^2 + 5 = -7x^2 + 3x + 5$$ ### Step 5: Final Expression $$f'(x) = \frac{(x - 1)^2 (x^2 + x)^4 (-7x^2 + 3x + 5)}{(x^2 + x)^{10}}$$ Cancel $$(x^2 + x)^4$$ from the numerator and denominator: $$f'(x) = \frac{(x - 1)^2 (-7x^2 + 3x + 5)}{(x^2 + x)^6}$$ --- Would you like a more detailed explanation of any step? Here are some related questions: 1. How do you apply the chain rule in differentiation? 2. What is the power rule in calculus? 3. How do you simplify expressions with common factors? 4. Can you explain how to use implicit differentiation? 5. What are some real-world applications of derivatives? **Tip:** The quotient rule is useful when dealing with fractions, but sometimes rewriting a function as a product can make differentiation easier!