Formal Proof of the Chain Rule?

Suppose h(x) = f o g(x)

We have by the definition of the derivative for g

g(x+e) = g(x) + e g'(x) + e o_e(1)

where o_e(1) is a function of e that -> 0 as e-> 0. Thus

f(g(x+e)) = f(g(x) + e g'(x) + e o_e(1))

Now, by the definition of the derivative of f, for d not equal to 0

f(g(x) + d) = f(g(x)) + d*f'(x) + d*o_d(1)

where once again o_d(1) is a function of d such that o(1) -> 0 as d -> 0. On the other hand, if d = 0, then

f(g(x) + d) = f(g(x)) + d*f'(x).

If we allow d to depend on e and be such that d(e) -> 0 as e->0, then the above says

f(g(x) + d(e)) = f(g(x)) + d(e)*f'(g(x)) + d(e)*o_{e,2}(1)

where o_{e,2}(1) is also a function of e that goes to 0 as e -> 0. Now of course if we set d(e) = e g'(x) + e o_e(1) then combining all of the above gives

f(g(x+e)) = f(g(x)) +(e g'(x) + e o_e(1))*f'(g(x)) + (e g'(x) + e o_e(1))*o_{e,2}(1).

Therefore

(f(g(x+e)) - f(g(x)))/e = (g'(x) + o_e(1))*f'(g(x)) + (g'(x) + o_e(1))*o_{e,2}(1).

Taking the limit e->0 on both sides gives

h'(x) = f'(g(x)) g'(x).

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