1. Formal derivative

For a polynomial $f(x)$ reduce $\frac {f(x)-f(z)}{x-z}$, substitute Z in for X. Show that this is $f'(x)$.

From calculus I know that as the lim x approaches z that is the derivative, but if I just substitute, how would I compute that? Thanks.

For a polynomial $f(x)$ reduce $\frac {(f(x)-f(z)}{x-z}$, substitute Z in for X. Show that this is $f'(x)$.

From calculus I know that as the lim x approaches z that is the derivative, but if I just substitute, how would I compute that? Thanks.
By linearity, you just need to study the case of $f(X)=X^k$, i.e. to consider $\frac{X^k-z^k}{X-z}$. This is given by the usual formula: $\frac{X^k-z^k}{X-z}=X^{k-1}+zX^{k-2}+\cdots+z^{k-2}X+z^{k-1}$. And now, substitute.

3. Originally Posted by Laurent
This is given by the usual formula: $\frac{X^k-z^k}{X-z}=X^{k-1}+zX^{k-2}+\cdots+z^{k-2}X+z^{k-1}$. And now, substitute.
Thanks for your help, but may I ask what formula did we use here?

4. Hello,
Thanks for your help, but may I ask what formula did we use here?
$\frac{X^k-z^k}{X-z}=\frac{X^k \left(1-\left(\tfrac zX\right)^k\right)}{X \left(1-\frac zX\right)}$

$=X^{k-1} \times \frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}$

You can recognize that this is the sum of a geometric sequence :
$\sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\frac{1-\left(\tfrac zX\right)^k}{1-\frac zX}$

Hence
$\frac{X^k-z^k}{X-z}=X^{k-1} \sum_{i=0}^{k-1} \left(\tfrac zX\right)^i=\sum_{i=0}^{k-1} z^i \cdot X^{(k-1)-i}$

and this gives the formula Laurent wrote.