Learn the chain rule, which differentiates by distinguishing between the “inner function” and the “outer function.”
Differentiation of Composite Functions (Chain Rule)
When y = f(g(x)): y' = f'(g(x)) × g'(x)
Remember it as “differentiate the outer function × differentiate the inner function.”
Example ①
Differentiate y = (x² + 3)⁵.
Answer:Let u = x²+3, so y = u⁵.
y' = 5u⁴ × 2x = 5(x²+3)⁴ × 2x = 10x(x²+3)⁴
Differentiate y = (x² + 3)⁵.
Answer:Let u = x²+3, so y = u⁵.
y' = 5u⁴ × 2x = 5(x²+3)⁴ × 2x = 10x(x²+3)⁴
Example ②
Differentiate y = (2x−1)³.
Answer:The derivative of the outer function u³ is 3u², and the derivative of the inner function 2x−1 is 2.
y' = 3(2x−1)² × 2 = 6(2x−1)²
Differentiate y = (2x−1)³.
Answer:The derivative of the outer function u³ is 3u², and the derivative of the inner function 2x−1 is 2.
y' = 3(2x−1)² × 2 = 6(2x−1)²
Example ③
Differentiate y = √(x²+1). (√u = u^(1/2))
Answer:y' = (1/2)(x²+1)^(−1/2) × 2x = x/√(x²+1)
Differentiate y = √(x²+1). (√u = u^(1/2))
Answer:y' = (1/2)(x²+1)^(−1/2) × 2x = x/√(x²+1)
Key Points
- Differentiate the outer function (leaving the inside unchanged) × differentiate the inner function
- The expression inside parentheses or a radical is the “inner function”
- For combined product, quotient, and composite problems, decide first which rule to use
Practice Problems
- Differentiate y = (3x+2)⁴.
- Differentiate y = (x²−2x+1)³.
- Differentiate y = 1/(x²+1). (Use the derivative of (u)⁻¹.)
Answers and Explanations
- Answer:12(3x+2)³
Explanation: Outer derivative 4u³ × inner derivative 3 = 12(3x+2)³. - Answer:3(x²−2x+1)²(2x−2) = 6(x−1)(x²−2x+1)² = 6(x−1)⁵
Explanation: Since x²−2x+1=(x−1)², the whole expression is (x−1)⁶. Its derivative is 6(x−1)⁵, which agrees. - Answer:−2x/(x²+1)²
Explanation: y=(x²+1)^(−1). y'=−(x²+1)^(−2)×2x.