Proving the Chain Rule

Real Analysis · Axiom Academy

EXAMPLE Proving the Chain Rule A rigorous proof of , handling the subtle case where g'(x_0)=0 . Let g be differentiable at x_0 and let f be differentiable at g(x_0) . Prove that the composite h(x) = f(g(x)) is differentiable at x_0 and that . A small nudge in x is scaled by g'(x) on the way to u , then scaled again by f'(u) on the way to y . The composite's rate is the product of the two rates — that is the chain rule. The proof below makes this precise, even when g'(x_0)=0 . Nicely done. You proved the chain rule the careful way — the way that survives the case the quick proof breaks on. The pieces that make it work: Error-function form: writing with encodes differentiability of f without ever dividing by . The subtle case: when g'(x_0)=0 , the error term still vanishes, because multiplies a bounded factor. Continuity is the bridge: g differentiable g continuous , which is what forces . Result: — the outer rate at g(x_0) times the inner rate at x_0 . (Check: .) Tracking an error term until it provably vanishes is a core move in real analysis — you will use it again for products, quotients, and the rules built on top of the chain rule.

This is the written version of the interactive lesson above. See the full Real Analysis course.