1. ## [SOLVED] Finding Derivatives

One Problem States:
1. F(x) = 1/x , a=2
By Using The definition of a derivative
do i have to multiply the denominator and numerator by a fraction? Im getting a bit lost here and its not a good sign...

there is another problem that says to use the alternate definition of a derivative where the numerator has f(x) - f(a), instead of (x+h). and to find the derivative at the point:

2. (with the same problem above.)
What Is The Difference and what accounts for that difference?

2. Hello
Originally Posted by >_<SHY_GUY>_<
One Problem States:
1. F(x) = 1/x , a=2
By Using The definition of a derivative
do i have to multiply the denominator and numerator by a fraction? Im getting a bit lost here and its not a good sign...
$\displaystyle \frac{f(a+h)-f(a)}{h}=\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac 12}{h} \quad (*)$
now, by noting that 2(h+2) is the common denominator, $\displaystyle \frac{1}{2+h}-\frac 12=\frac{2}{2(2+h)}-\frac{2+h}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}$

$\displaystyle (*)=\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2h(2+h)}=-\frac{1}{2(2+h)}$

now take the limit h -> 0 and you'll have the answer

there is another problem that says to use the alternate definition of a derivative where the numerator has f(x) - f(a), instead of (x+h). and to find the derivative at the point:

2. (with the same problem above.)
What Is The Difference and what accounts for that difference?
There is no difference

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

from the first one to the second one, you just have to substitute $\displaystyle x=a+h$

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

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

4. Originally Posted by Moo
Hello

$\displaystyle \frac{f(a+h)-f(a)}{h}=\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac 12}{h} \quad (*)$
now, by noting that 2(h+2) is the common denominator, $\displaystyle \frac{1}{2+h}-\frac 12=\frac{2}{2(2+h)}-\frac{2+h}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}$

$\displaystyle (*)=\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2h(2+h)}=-\frac{1}{2(2+h)}$

now take the limit h -> 0 and you'll have the answer

There is no difference

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

from the first one to the second one, you just have to substitute $\displaystyle x=a+h$
Hola

Ok, I Think i was getting confused with substituting a in finding the derivative (sounds weird, butt its how i word it)
so i substitute what a is in f(a+h) and make it similar to what f(x), and f(a) is just substituting. ok i get it
Gracias

Sorry, Wrong Problem :[, but yes you are right

5. Originally Posted by >_<SHY_GUY>_<
Hola

Ok, I Think i was getting confused with substituting a in finding the derivative (sounds weird, butt its how i word it)
so i substitute what a is in f(a+h) and make it similar to what f(x), and f(a) is just substituting. ok i get it
Gracias

but isn't there a difference? i tried #2 and i get 0.
and the first one is 4.
Hmmm the first one is -1/4 !

and for #2, use the same method for $\displaystyle \frac 1x-\frac 12$
you should get the same things.