# Thread: Confused by alternate method of finding f'(a)

1. ## Confused by alternate method of finding f'(a)

I'm having a hard time using:

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

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

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

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

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

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

$=\lim_{x \to a}\dfrac{x^2-8x+9-a^2-8a+9}{x-a}$

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

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

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?

2. You want to write the numerator as $x^2-a^2-8(x-a)$. Then you can factor out $x-a$ and cancel with the denominator.

3. Originally Posted by Latvija13
I'm having a hard time using:

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

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

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

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

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

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

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

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

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

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?
That step should be $\displaystyle \lim_{x \to a}\frac{x^2 - 8x + 9 - a^2 + 8a - 9}{x - a}$, since you are subtracting EVERYTHING in that bracket...

4. Originally Posted by DrSteve
You want to write the numerator as $x^2-a^2-8(x-a)$. Then you can factor out $x-a$ and cancel with the denominator.
Don't I end up with just $f'(a) = -8$ doing that?

$=\lim_{x \to a}\dfrac{x^2-8x+9-a^2+8a-9}{x-a}$

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

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

I'm not sure how I get -8(x+a) to turn into -8(x-a) But to continue.

$=\lim_{x \to a}\dfrac{x^2-a^2-8(x-a)}{x-a}$

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

$=a^2-a^2-8$

$=-8$

Hate being bad at math :/

Edit: Actually, I think it was my subtraction error that got me.

Edit: I think I see how its done now.

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

I'm not sure how I get -8(x+a) to turn into -8(x-a) But to continue.
Edit: I see now, $-8(x-a)$ has be like that in order to get a positive a.

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

Cancel out my two $(x-a)$'s

Then I end up with:

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

$={(a+a)-8$

$=2a-8$

5. Your computation after the last edit is correct.