Proposition 6.2.1. the Derivative is Linear.
If \(f,g\) are differentiable at \(c\) and \(\lambda\in \mathbb{R}\text{,}\) then \(\lambda f+g\) is differentiable at \(c\text{;}\) moreover, \(D(\lambda f+g)(c)=\lambda Df(c)+Dg(c)\text{.}\)
Section 6.2 Derivative Rules
This won’t take long.
Theorem 6.2.2. Leibniz’s Rule.
If \(f,g:\mathbb{R}\to\mathbb{R}\) are both differentiable at \(c\text{,}\) then so is \(fg\text{;}\) moreover, \((fg)'(c)=f'(c)g(c)+f(c)g'(c)\text{.}\)
Proof.
Here’s the Carathéodory version.
We know there are functions \(\Phi\text{,}\) \(\Psi\text{,}\) both continuous at \(c\text{,}\) so that
\begin{equation*}
f(x)=f(c)+\Phi(x)(x-c)
\end{equation*}
and
\begin{equation*}
g(x)=g(c)+\Psi(x)(x-c)\ \ .
\end{equation*}
So we have
\begin{align*}
f(x)g(x)&=\left(f(c)+\Phi(x)(x-c)\right)\left(g(c)+\Psi(x)(x-c)\right)\\
& f(c)g(c)+\Phi(x)(x-c)g(c)+f(c)\Psi(x)(x-c)+\Phi(x)\Psi(x)(x-c)^2
\end{align*}
so we can take the function \(\Theta(x)=\Phi(x)g(c)+f(c)\Psi(x)+\Phi(x)\Psi(x)(x-c)\) to see that \(fg\) is differentiable according to Carathéodory.
Evaluating, we get \(\Theta(c)=\Phi(c)g(c)+f(c)\Psi(c)=f'(c)g(c)+f(c)g'(c)\text{.}\)
Theorem 6.2.3. The Chain Rule.
If \(f,g\) are differentiable and defined so that the composition \(g\circ f\) makes sense, then \(g\circ f\) is differentiable; moreover, \(D(g\circ f)(c)=\left((Dg)(f(c))\right)(Df(c))\text{.}\)
Proof.
Again, let’s use Carathéodory.
We know that there are \(\Phi\) so that \(f(x)=f(c)+\Phi(x)(x-c)\) and \(\Psi\) so that \(g(y)=g(f(c))+\Psi(y)(y-f(c))\text{.}\) So we have
\begin{align*}
g(f(x))&=g(f(c))+\Psi(f(x))(f(x)-f(c))\\
&g(f(c))+\Psi(f(x))\left(\Psi(x)(x-c)\right)
\end{align*}
So that setting \(\Theta(x)=\Psi(f(x))\Psi(x)\text{,}\) we have
\begin{equation*}
g(f(x))=g(f(c))+\Theta(x)(x-c)\ \ .
\end{equation*}
Since \(\Phi\) is continuous at \(c\) and \(\Psi\) is continuous at \(f(c)\text{,}\) we get that \(\Theta\) is continuous at \(c\text{.}\) Thus \(g\circ f\) is differentiable according to Carathéodory at \(c\text{.}\)
Remark 6.2.4.
One reason I personally prefer the Carathéodory approach is that we don’t need to know what the formula for the claimed derivative is going in -- notice that the actual derivative rule just falls out at the end by plugging in \(c\text{.}\) To use the Fréchet approach, you have to know the formula going in.
Checkpoint 6.2.5.
Try your hand at proving Leibniz’s Rule and the Chain Rule using the Fréchet version: show that
\begin{equation*}
\displaystyle\lim_{x\to c} \frac{f(x)g(x)-f(c)g(c)-\left(Df(c)g(c)-f(c)Dg(c)\right)(x-c)}{\lVert x-c\rVert}=0
\end{equation*}
and
\begin{equation*}
\displaystyle\lim_{x\to c} \frac{g(f(x))-g(f(c))-\left((Dg)(f(c))Df(c)\right)(x-c)}{\lVert x-c\rVert}=0\ \ .
\end{equation*}
