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Differentiation Rules

Once you understand what a derivative means conceptually, differentiation rules let you calculate derivatives quickly without going back to the limit definition every time. These four rules handle the vast majority of functions you will encounter.

In this lesson

  1. The power rule
  2. The product rule
  3. The quotient rule
  4. The chain rule
  5. Combining the rules

1 The Power Rule

For f(x) = xⁿ: f'(x) = nxⁿ⁻¹. Bring the exponent down as a coefficient, reduce the exponent by 1.

Power Rule Examples
Differentiate each function
1f(x) = x⁵ → f'(x) = 5x⁴
2f(x) = x² → f'(x) = 2x
3f(x) = x → f'(x) = 1 (since x = x¹)
4f(x) = 1 = x⁰ → f'(x) = 0 (constants have zero derivative)
5f(x) = √x = x^(1/2) → f'(x) = (1/2)x^(−1/2) = 1/(2√x)
6f(x) = 1/x = x^(−1) → f'(x) = −x^(−2) = −1/x²
Answer: The power rule handles any real exponent — integer, fraction, or negative

For sums and differences: differentiate term by term. d/dx[f(x) + g(x)] = f'(x) + g'(x). For constants multiplied by functions: d/dx[cf(x)] = c·f'(x).

2 The Product Rule

For f(x) = u(x)·v(x): f'(x) = u'v + uv'

"First times derivative of second, plus second times derivative of first." The derivative of a product is NOT the product of the derivatives.

Product Rule
Differentiate f(x) = x²·sin(x)
1Let u = x², v = sin(x)
2u' = 2x, v' = cos(x)
3f'(x) = u'v + uv' = 2x·sin(x) + x²·cos(x)
Answer: f'(x) = 2x sin(x) + x² cos(x)
Don't multiply derivatives

d/dx[x² · x³] ≠ 2x · 3x². The correct answer uses the product rule (or just simplify first: x⁵ → 5x⁴). Always simplify first if possible.

3 The Quotient Rule

For f(x) = u(x)/v(x): f'(x) = (u'v − uv') / v²

Memory trick: "Low d-High minus High d-Low, all over Low squared." (Low = denominator, High = numerator, d = derivative)

Quotient Rule
Differentiate f(x) = x²/(x+1)
1u = x², v = x+1
2u' = 2x, v' = 1
3f'(x) = (u'v − uv') / v² = (2x(x+1) − x²(1)) / (x+1)²
4= (2x² + 2x − x²) / (x+1)² = (x² + 2x) / (x+1)²
5= x(x+2) / (x+1)²
Answer: f'(x) = x(x+2)/(x+1)²

4 The Chain Rule

For composite functions f(g(x)): d/dx[f(g(x))] = f'(g(x)) · g'(x)

"Derivative of the outside (keeping inside unchanged), times derivative of the inside."

Chain Rule
Differentiate f(x) = (x² + 3)⁵
1Outer function: u⁵, inner function: u = x² + 3
2Derivative of outer: 5u⁴ = 5(x²+3)⁴
3Derivative of inner: 2x
4Chain rule: f'(x) = 5(x²+3)⁴ · 2x = 10x(x²+3)⁴
Answer: f'(x) = 10x(x²+3)⁴
Recognizing chain rule

Any time you see a function raised to a power, a trig function of an expression, or a composition of functions, the chain rule applies. The signal: 'there's something inside something else.'

5 Combining the Rules

Real problems often require multiple rules together. The key is identifying the structure: what's the outermost operation? Apply that rule first, then handle the inner parts.

Combined Rules
Differentiate f(x) = x³ · (x²+1)⁴
1This is a product of x³ and (x²+1)⁴ — use product rule
2u = x³, v = (x²+1)⁴
3u' = 3x², v' = 4(x²+1)³ · 2x = 8x(x²+1)³ (chain rule on v)
4f'(x) = 3x²(x²+1)⁴ + x³ · 8x(x²+1)³
5= 3x²(x²+1)⁴ + 8x⁴(x²+1)³
6Factor: x²(x²+1)³[3(x²+1) + 8x²] = x²(x²+1)³(11x²+3)
Answer: f'(x) = x²(x²+1)³(11x²+3)

Practice Problems

Differentiate f(x) = 3x⁴ − 2x² + 5x − 7
Power rule term by term: 12x³ − 4x + 5
Differentiate f(x) = (2x + 1)³ using the chain rule
Outer: u³ → 3u². Inner: 2x+1 → 2. Result: 3(2x+1)² · 2 = 6(2x+1)²
Differentiate f(x) = x/(x²+1) using the quotient rule
u=x, v=x²+1, u'=1, v'=2x. f'=(1·(x²+1) − x·2x)/(x²+1)² = (x²+1−2x²)/(x²+1)² = (1−x²)/(x²+1)²