User:IssaRice/Chain rule proofs

Using Newton's approximation

Since $g$ is differentiable at $y_0$, we know $g^{\prime }(y_0)$ is a real number, and we can write

$g(y)=g(y_0)+g^{\prime }(y_0)(y-y_0)+[g(y)-(g(y_0)+g^{\prime }(y_0)(y-y_0))]$ (there is no magic: the terms just cancel out)

If we define $E_g(\mathrm{\Delta }y):=g(y)-(g(y_0)+g^{\prime }(y_0)(y-y_0))$ we can write

$g(y)=g(y_0)+g^{\prime }(f(x_0))(y-y_0)+E_g(\mathrm{\Delta }y)$

Newton's approximation says that $|E_g(\mathrm{\Delta }y)|\le ϵ|y-y_0|$ as long as $|y-y_0|\le \delta$.

Since $f$ is differentiable at $x_0$, we know that it must be continuous at $x_0$. This means we can keep $|f(x)-y_0|\le \delta$ as long as we keep $|x-x_0|\le {\delta }^{\prime }$.

Since $f(x)\in Y$ and $|f(x)-y_0|\le \delta$, this means we can substitute $y=f(x)$ and get

$g(f(x))=g(y_0)+g^{\prime }(f(x_0))(f(x)-y_0)+E_g(\mathrm{\Delta }f)$

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

$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+[f(x)-(f(x_0)+f^{\prime }(x_0)(x-x_0))]$

Again, we can define $E_f(\mathrm{\Delta }x):=f(x)-(f(x_0)+f^{\prime }(x_0)(x-x_0))$ and write this as

$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+E_f(\mathrm{\Delta }x)$

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

$g(f(x))=g(y_0)+g^{\prime }(f(x_0))(f^{\prime }(x_0)(x-x_0)+E_f(\mathrm{\Delta }x))+E_g(\mathrm{\Delta }f)$

where we have canceled out two terms using $f(x_0)=y_0$.

Thus we have

$g(f(x))=g(y_0)+g^{\prime }(f(x_0))f^{\prime }(x_0)(x-x_0)+[g^{\prime }(f(x_0))E_f(\mathrm{\Delta }x)+E_g(\mathrm{\Delta }f)]$

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 $|x-x_0|$.