# Math Help - irritating derivative problem

1. ## irritating derivative problem [solved]

hi, i have a test on derivatives tomorrow and was going over some problems when I got stumped. I have the problem and the answer, but i want to know how to get the answer. thanks

use definition fprime (a) = (lim as h approaches 0) [f(a+h) - f(a)] / h

to find the derivative of the given function at the indicated point

f(x) = 1/x, a = 2

nvm, brain cramp, i figured it out

2. ## Re: irritating derivative problem

$f'(a) = \lim_{h \to 0} \frac{\frac{1}{a+h} - \frac{1}{a}}{h}$

$= \lim_{h \to 0} \frac{\frac{a}{a(a+h)} - \frac{a+h}{a(a+h)}}{h}$

$= \lim_{h \to 0}\left(\frac{1}{h}\right)\left(\frac{-h}{a(a+h)}\right)$

$= \lim_{h \to 0}\frac{-1}{a(a+h)}$

$= \frac{-1}{a^2}$, provided a ≠ 0.

since 2 ≠ 0, f'(2) = -1/4.

3. ## Re: irritating derivative problem

Originally Posted by TacticalPro
hi, i have a test on derivatives tomorrow and was going over some problems when I got stumped. I have the problem and the answer, but i want to know how to get the answer. thanks

use definition fprime (a) = (lim as h approaches 0) [f(a+h) - f(a)] / h

to find the derivative of the given function at the indicated point

f(x) = 1/x, a = 2
$\lim_{h \to 0} \frac{1}{h} \left(\frac{1}{2+h} - \frac{1}{2}\right)$

get a common denominator and combine the fractions in ( ) ... do it right and you'll get that h in the denominator to cancel out ... then determine the limit.

4. ## Re: irritating derivative problem

Originally Posted by skeeter
$\lim_{h \to 0} \frac{1}{h} \left(\frac{1}{2+h} - \frac{1}{2}\right)$

get a common denominator and combine the fractions in ( ) ... do it right and you'll get that h in the denominator to cancel out ... then determine the limit.
thanks, when i first looked at the problem i completely forgot about rearranging to get the h out of the botom

5. ## Re: irritating derivative problem

Originally Posted by Deveno
$\lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

$= \lim_{h \to 0} \frac{\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)}}{h}$

$= \lim_{h \to 0}\left(\frac{1}{h}\right)\left(\frac{-h}{x(x+h)}\right)$

$= \lim_{h \to 0}\frac{-1}{x(x+h)}$

$= \frac{-1}{x^2}$, provided x ≠ 0.

since 2 ≠ 0, f'(2) = -1/4.
thanks!!!! I actually plugged in for x before rearranging and I find I like it your way better: rearranging the equation in terms of x, and then putting in 2 for x to get the derivative. !!! yay