This section provides comprehensive practice with the differentiation of products, quotients, and composite functions, as well as application problems.
Example ①
Differentiate y = x(x−1)³.
Answer:f=x, g=(x−1)³. f'=1, g'=3(x−1)².
y' = (x−1)³ + x × 3(x−1)² = (x−1)²((x−1)+3x) = (x−1)²(4x−1)
Differentiate y = x(x−1)³.
Answer:f=x, g=(x−1)³. f'=1, g'=3(x−1)².
y' = (x−1)³ + x × 3(x−1)² = (x−1)²((x−1)+3x) = (x−1)²(4x−1)
Example ②
Find the extreme values of y = (x+1)²(x−2)².
Answer:y' = 2(x+1)(x−2)² + (x+1)² × 2(x−2) = 2(x+1)(x−2)(x−2 + x+1) = 2(x+1)(x−2)(2x−1).
There are sign changes at x = −1, 1/2, 2.
x=−1: local maximum y=0 ←y=0 (actually determined by continuity)→ y(−1)=0, x=1/2: local minimum y=(3/2)²(−3/2)²=81/16, x=2: local maximum y=0.
Find the extreme values of y = (x+1)²(x−2)².
Answer:y' = 2(x+1)(x−2)² + (x+1)² × 2(x−2) = 2(x+1)(x−2)(x−2 + x+1) = 2(x+1)(x−2)(2x−1).
There are sign changes at x = −1, 1/2, 2.
x=−1: local maximum y=0 ←y=0 (actually determined by continuity)→ y(−1)=0, x=1/2: local minimum y=(3/2)²(−3/2)²=81/16, x=2: local maximum y=0.
Review
- Product: (fg)' = f'g + fg'
- Quotient: (f/g)' = (f'g − fg')/g²
- Composite function: (f∘g)' = f'(g) × g'
- Proof of an inequality: Treat the difference as a function and find its minimum value
Practice Problems
- Differentiate y = (x²+1)/(x−1).
- Find the extreme values of y = (x+2)²(x−1).
- Differentiate y = x√(x+1).
Answers and Explanations
- Answer:(x²−2x−1)/(x−1)²
Explanation: Quotient rule. (2x(x−1)−(x²+1))/(x−1)²=(x²−2x−1)/(x−1)². - Answer:f'=2(x+2)(x−1)+(x+2)²=(x+2)(2(x−1)+(x+2))=(x+2)(3x)=3x(x+2). At x=−2, local maximum f(−2)=0; at x=0, local minimum f(0)=−4.
Explanation: Sign: f'>0 for x<−2, f'<0 for −2 - Answer:√(x+1) + x/(2√(x+1)) = (2(x+1)+x)/(2√(x+1)) = (3x+2)/(2√(x+1))
Explanation: y=x(x+1)^(1/2). y'=(x+1)^(1/2)+x(1/2)(x+1)^(−1/2). Combine the fractions and simplify.