The first derivative of Taylor's polynomial

Question

When I apply L'Hopital's rule n times to prove Taylor's Theorem, it comes out that the first derivative of Taylor's polynomial is $P_{n-1}f'(x;x_0)$:

$$ \begin{split} \lim_{x\to x_0} \frac{f(x)-P_nf(x;x_0)}{(x-x_0)^n} &= \lim_{x\to x_0} \frac{f'(x)-P_{n-1}f'(x;x_0)}{n(x-x_0)^{n-1}} \\ &= \ldots \\ &= \lim_{x\to x_0}\frac{f^{(n)}(x)-f^{(n)}(x_0)}{n!}=0 \end{split} $$

How can I show that $$\frac{d}{dx}P_nf(x;x_0) = P_{n-1}f'(x;x_0)?$$

Answer 1

You've defined (in the comments) that

$$P_nf(x;x_0)=\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

Taking the derivative, you should end up with

$$\frac d{dx}P_nf(x;x_0)=\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}k(x-x_0)^{k-1}$$

Notice that $f^{(k)}(x_0)$ is a constant.

When $k=0$, the summand is $0$, so we can ignore that. Simplifying a bit, we get

$$\frac d{dx}P_nf(x;x_0)=\sum_{k=1}^n \frac{f^{(k)}(x_0)}{(k-1)!}(x-x_0)^{k-1}=\sum_{k=0}^{n-1} \frac{f^{(k+1)}(x_0)}{k!}(x-x_0)^k$$

which by definition, is equal to $P_{n-1}f'(x;x_0)$