Proof of the Chain Rule

Statement

Let g be differentiable at x and let f be differentiable at g(x). Then the composite function F(x) = f(g(x)) is differentiable at x and:

In Leibniz notation, if y = f(u) and u = g(x), then:

Proof

Set u = g(x) and write the difference quotient for F(x) = f(g(x)):

Let Δu = g(x+h) − g(x), so g(x+h) = g(x) + Δu = u + Δu. Note that Δu → 0 as h → 0 (because g is differentiable, hence continuous, at x).

Rewrite the numerator using Δu:

Multiply and divide by Δu (valid whenever Δu ≠ 0):

As h → 0, the first factor tends to f′(u) = f′(g(x)) (by the definition of the derivative of f at u), and the second factor Δu/h = [g(x+h)−g(x)]/h tends to g′(x). Therefore:

Note: The argument above is valid when Δu ≠ 0 for small h. A more careful proof handles the case Δu = 0 by defining an auxiliary function that removes the division; the conclusion is the same.

Common Mistakes

• Forgetting the inner derivative: d/dx cos(5x) = −sin(5x)·5, not just −sin(5x). The factor 5 comes from d/dx (5x) = 5.
• Evaluating the outer derivative at x instead of g(x): For d/dx (x²+1)³, the outer derivative is 3u² evaluated at u = x²+1, giving 3(x²+1)². The inner derivative is 2x. Result: 6x(x²+1)².
• Confusing order of composition: f(g(x)) and g(f(x)) are different. Always identify which function is innermost.