Originally Posted by

**Latvija13** I'm having a hard time using:

$\displaystyle f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}$

to find the derivative of $\displaystyle f(x)=x^2-8x+9$

I need to do it using $\displaystyle f'(a)=\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}$

and the other way. I did fine finding it with the second method and came up with $\displaystyle f'(a) = 2a-8$

Here is the order that I've tried this with the first method.

$\displaystyle f'(a)=\lim_{x \to a}\dfrac{x^2-8x+9-(a^2-8a+9)}{x-a}$

$\displaystyle =\lim_{x \to a}\dfrac{x^2-8x+9-a^2-8a+9}{x-a}$ Your mistake is here

$\displaystyle =\lim_{x \to a}\dfrac{x^2-8x-a^2+8a}{x-a}$

$\displaystyle =\lim_{x \to a}\dfrac{x(x-8)-a(a+8)}{x-a}$

I'm not sure what to do at this point. Did I mess up and can some how use factor by grouping and then cancel out (x-a) and use substitution to get my derivative?