Thread: The definition of a derivative

$\displaystyle f(x)=\frac{3}{2x+1}$

I got

the limit = $\displaystyle 3(\frac{-2}{ (2x+1)^2}) $

When you use the definition, the limit will involve (depending on the form you use), the letter h, a, x_0, or delta x.

You should have
$\displaystyle \lim_{x \to h}\frac{f(x + h) - f(x)}{h}$

$\displaystyle \lim_{x \to h}\frac{\frac{3}{2(x + h) + 1} - \frac{3}{2x + 1}}{h}$

for starters, at least.

I have that, I'm trying to check my answer and that is where I got stuck I would type it all out but i'm extremely tired

Originally Posted by purplec16

I have that, I'm trying to check my answer and that is where I got stuck I would type it all out but i'm extremely tired

For f(x) = 3/(2x + 1)
f'(x) = -6/(2x + 1)^2 , which is what you have (after simplifying).

You can check using wolfram... y = 3/(2x + 1) - Wolfram|Alpha