User:IssaRice/Chain rule proofs

Using Newton's approximation

Since $g$ is differentiable at $y_{0}$ , we know $g^{'} (y_{0})$ is a real number, and we can write

$g (y) = g (y_{0}) + g^{'} (y_{0}) (y - y_{0}) + [g (y) - (g (y_{0}) + g^{'} (y_{0}) (y - y_{0}))]$ (there is no magic: the terms just cancel out)

If we define $E_{g} (Δ y) : = g (y) - (g (y_{0}) + g^{'} (y_{0}) (y - y_{0}))$ we can write

$g (y) = g (y_{0}) + g^{'} (f (x_{0})) (y - y_{0}) + E_{g} (Δ y)$

Newton's approximation says that $| E_{g} (Δ y) | \leq ϵ | y - y_{0} |$ as long as $| y - y_{0} | \leq δ$ .

Since $f$ is differentiable at $x_{0}$ , we know that it must be continuous at $x_{0}$ . This means we can keep $| f (x) - y_{0} | \leq δ$ as long as we keep $| x - x_{0} | \leq δ^{'}$ .

Since $f (x) \in Y$ and $| f (x) - y_{0} | \leq δ$ , this means we can substitute $y = f (x)$ and get

$g (f (x)) = g (y_{0}) + g^{'} (f (x_{0})) (f (x) - y_{0}) + E_{g} (Δ f)$

Now we use the differentiability of $f$ . We can write

$f (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + [f (x) - (f (x_{0}) + f^{'} (x_{0}) (x - x_{0}))]$

Again, we can define $E_{f} (Δ x) : = f (x) - (f (x_{0}) + f^{'} (x_{0}) (x - x_{0}))$ and write this as

$f (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + E_{f} (Δ x)$

Now we can substitute this into the expression for $g (f (x))$ to get

$g (f (x)) = g (y_{0}) + g^{'} (f (x_{0})) (f^{'} (x_{0}) (x - x_{0}) + E_{f} (Δ x)) + E_{g} (Δ f)$

where we have canceled out two terms using $f (x_{0}) = y_{0}$ .

Thus we have

$g (f (x)) = g (y_{0}) + g^{'} (f (x_{0})) f^{'} (x_{0}) (x - x_{0}) + [g^{'} (f (x_{0})) E_{f} (Δ x) + E_{g} (Δ f)]$

The stuff in square brackets is our "error term" for $g \circ f$ . Now we just need to make sure it is small, even after dividing by $| x - x_{0} |$ .