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- I was wondering if someone could look over my proof of the chain rule that I wrote up while following the guidelines out of Taylor and Mann's Advanced Calculus.

I am currently self-studying Taylor and Mann's Advanced Calculus (3rd edition, specifically). I stumbled across their guidelines for a proof of the chain rule, leaving the rest of the proof up to the reader to complete.

I was wondering if someone could look over my proof, and point out any flaws I might have made.

Theorem:

Suppose g is differentiable at a point t0 of the interval α<t

Proof:

Consider any nonzero value of Δt so small that α<t0+Δt

Δx = g(t0 + Δt) - g(t0),

Δy = f(x0 + Δx) - f(x0),

ε = Δy/Δx - f'(x0).

Notice that

F(t0 + Δt) - F(t0) / Δt

= f(x0 + Δx) - f(x0) / Δt

= Δy/Δt

= Δx(Δy/Δx - f'(x0) + f'(x0)) / Δt

= Δx(ε + f'(x0)) / Δt.

Then,

limΔt->0 [F(t0 + Δt) - F(t0) / Δt]

= limΔt->0[g(t0 + Δt) - g(t0) / Δt] * limΔt->0[ε + f'(x0)]. (1)

Since limΔt->0[Δx] = g(t0) - g(t0) = 0, limΔx->0[ε] = f'(x0) - f'(x0) = 0. So, Δx->0 and ε->0 as Δt->0. (1) becomes

F'(t0) = g'(t0)f'(x0).

I was wondering if someone could look over my proof, and point out any flaws I might have made.

Theorem:

Suppose g is differentiable at a point t0 of the interval α<t

**<**β. Let x0 = g(t0), and suppose that f is differentiable at x0. Then the composite function F(x0) is differentiable at t0, and F'(t0) = f'(x0)g'(t0).Proof:

Consider any nonzero value of Δt so small that α<t0+Δt

**<**β and defineΔx = g(t0 + Δt) - g(t0),

Δy = f(x0 + Δx) - f(x0),

ε = Δy/Δx - f'(x0).

Notice that

F(t0 + Δt) - F(t0) / Δt

= f(x0 + Δx) - f(x0) / Δt

= Δy/Δt

= Δx(Δy/Δx - f'(x0) + f'(x0)) / Δt

= Δx(ε + f'(x0)) / Δt.

Then,

limΔt->0 [F(t0 + Δt) - F(t0) / Δt]

= limΔt->0[g(t0 + Δt) - g(t0) / Δt] * limΔt->0[ε + f'(x0)]. (1)

Since limΔt->0[Δx] = g(t0) - g(t0) = 0, limΔx->0[ε] = f'(x0) - f'(x0) = 0. So, Δx->0 and ε->0 as Δt->0. (1) becomes

F'(t0) = g'(t0)f'(x0).