Chain Rule Formula and Derivation
Bikash Roy
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Full bio →The Chain Rule
If y = f(g(x)), then:
Leibniz notation:
dy/dx = (dy/du) × (du/dx) where u = g(x)Prime notation:
[f(g(x))]' = f'(g(x)) · g'(x)Both express the same rule: differentiate the outer function (evaluated at the inner), then multiply by the derivative of the inner function.
Why It Works
The chain rule follows from the limit definition of the derivative. For a composite function, the rate of change of y with respect to x decomposes into the rate y changes with u, times the rate u changes with x:
dy/dx = lim[Δx→0] Δy/Δx = lim (Δy/Δu) × (Δu/Δx) = (dy/du) × (du/dx)(The formal proof handles the case where Δu = 0 separately, but the intuition holds.)
Common Derivatives via Chain Rule
| Function | Let u = | dy/du | du/dx | Result |
|---|---|---|---|---|
| sin(x²) | x² | cos(u) | 2x | 2x·cos(x²) |
| e^(3x) | 3x | e^u | 3 | 3e^(3x) |
| ln(sin x) | sin x | 1/u | cos x | cos(x)/sin(x) = cot(x) |
| (x²+1)^5 | x²+1 | 5u^4 | 2x | 10x(x²+1)^4 |
| √(2x+1) | 2x+1 | 1/(2√u) | 2 | 1/√(2x+1) |
| cos(e^x) | e^x | −sin(u) | e^x | −e^x·sin(e^x) |
Multiple Chain Rule Applications
For triple compositions y = f(g(h(x))):
dy/dx = f'(g(h(x))) · g'(h(x)) · h'(x)Work from outside in, multiplying derivatives at each layer.
Example: y = sin(e^(x²)):
- Outer: f(u) = sin(u) → f’(u) = cos(u)
- Middle: g(v) = e^v → g’(v) = e^v
- Inner: h(x) = x² → h’(x) = 2x
- Result: cos(e^(x²)) · e^(x²) · 2x
Combined with Product and Quotient Rules
Chain rule often combines with other rules. For y = x²·sin(x³):
- Product rule: (x²)’ · sin(x³) + x² · [sin(x³)]’
- Chain rule on sin(x³): cos(x³) · 3x²
- Result: 2x·sin(x³) + x²·3x²·cos(x³) = 2x·sin(x³) + 3x⁴·cos(x³)
For a complete organized reference of every chain rule pattern by function type (trig, exponential, log, inverse trig, power), see the Chain Rule Patterns Reference. To practice applying this formula across eight problems of increasing difficulty, see the Chain Rule Practice Problems guide.