Fix the mistake: Differentiation ② (Product Rule, Composite Functions, Tables of Variation, and Extrema)

Differentiation of Products and Quotients

Use special formulas to differentiate products and quotients that cannot be expanded.

Product Rule

(f·g)' = f'g + fg'

Quotient Rule

(f/g)' = (f'g − fg')/g²

Example ①
Differentiate y = (x² + 1)(x³ − 2x) using the product rule.
Answer:f = x²+1, g = x³−2x. f' = 2x, g' = 3x²−2.
y' = 2x(x³−2x) + (x²+1)(3x²−2)
= 2x⁴−4x² + 3x⁴+x²−6 = 5x⁴−3x²−6

Example ②
Differentiate y = x²/(x+1) using the quotient rule.
Answer:f = x², g = x+1. f' = 2x, g' = 1.
y' = (2x(x+1) − x²) / (x+1)² = (x² + 2x)/(x+1)² = x(x+2)/(x+1)²

Key Points

Product: f'g + fg' (differentiate the first times the second + the first times the derivative of the second)
Quotient: (f'g − fg')/g² (subtract using the derivative of the denominator)
When expansion is possible, expanding may sometimes be faster

Practice Problems

Differentiate y = (x+1)(x²−x+1) using the product rule. (Also check by expanding.)
Differentiate y = (x²+3)/(x−1).
Differentiate y = (2x−1)(x³+x).

Answers and Explanations

Answer:y' = (x²−x+1) + (x+1)(2x−1) = x²−x+1+2x²+x−1 = 3x²
Explanation: Expanding gives x³+1, so y'=3x², which agrees.
Answer:(x²−2x−3)/(x−1)²
Explanation: (2x(x−1)−(x²+3))/(x−1)² = (2x²−2x−x²−3)/(x−1)².
Answer:2(x³+x)+(2x−1)(3x²+1) = 2x³+2x+6x³+2x−3x²−1 = 8x³−3x²+4x−1
Explanation: f'=2, g'=3x²+1. Apply the product rule.