# Definition of the Derivative

• Nov 27th 2010, 11:58 AM
wair
Definition of the Derivative
Using the definition of the derivative, find the derivative of f(x) = 1/(x-3)^2

ok so first I am using the equation f(x+h)-f(x)/h

I start out with (1/(x-3)^2 +h)-1/(x-3)^2 all of that over h. then i get a common denominator for (1/(x-3)^2 +h). Then I realize something is definitely wrong. Please help me.
• Nov 27th 2010, 12:11 PM
Quote:

Originally Posted by wair
Using the definition of the derivative, find the derivative of f(x) = 1/(x-3)^2

ok so first I am using the equation f(x+h)-f(x)/h

I start out with (1/(x-3)^2 +h)-1/(x-3)^2 all of that over h. then i get a common denominator for (1/(x-3)^2 +h). Then I realize something is definitely wrong. Please help me.

Replace x with x+h.

You are not bringing in the h with x, it's outside the brackets!

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

$f(x+h)=\displaystyle\frac{1}{(x+h-3)^2}$
• Nov 27th 2010, 12:13 PM
skeeter
Quote:

Originally Posted by wair
Using the definition of the derivative, find the derivative of f(x) = 1/(x-3)^2

ok so first I am using the equation f(x+h)-f(x)/h

I start out with (1/(x-3)^2 +h)-1/(x-3)^2 all of that over h. then i get a common denominator for (1/(x-3)^2 +h). Then I realize something is definitely wrong. Please help me.

$\displaystyle \frac{1}{h} \left[\frac{1}{[(x+h)-3]^2} - \frac{1}{(x-3)^2}\right]
$

$\displaystyle \frac{1}{h} \left[\frac{(x-3)^2 - [(x+h)-3]^2}{[(x+h)-3]^2(x-3)^2}\right]$

$\displaystyle \frac{1}{h} \left[\frac{(x^2-6x+9) - [(x+h)^2 - 6(x+h)+ 9]}{[(x+h)-3]^2(x-3)^2}\right]$

now ... expand the numerator in the [...], combine like terms, and factor out an h from what's left ... once done, you should be able to take the limit as $h \to 0$
• Nov 27th 2010, 01:01 PM
wair
Oh I see so I replace the x by x+h thank you.
• Nov 27th 2010, 01:46 PM
wair
After solving everything i get, -10x +24/ 9x^4-54x^3+81x^2. can someone confirm if this is correct ?
• Nov 27th 2010, 01:53 PM
skeeter
Quote:

Originally Posted by wair
After solving everything i get, -14x +12/ 9x^4-54x^3+81x^2. can someone confirm if this is correct ?

you should have got ...

$\displaystyle f'(x) = -\frac{2}{(x-3)^3}$ ,

... correct?
• Nov 27th 2010, 01:58 PM
wair
hmm well when I expanded numerator in [..] i got x^2+2xh+h^2-6x-6h+9. then i added and canceled terms i got -12x+18+2xh+h^2-6h. i factored out an h, -12x+18+h(2x+h+6). the i canceled the outer 1/h with the factored h and placed zero for all the remaining h's.
• Nov 27th 2010, 02:10 PM
skeeter
Quote:

Originally Posted by wair
hmm well when I expanded numerator in [..] i got x^2+2xh+h^2-6x-6h+9. then i added and canceled terms i got -12x+18+2xh+h^2-6h. i factored out an h, -12x+18+h(2x+h+6). the i canceled the outer 1/h with the factored h and placed zero for all the remaining h's.

you really need to be more careful with your signs.

redoing the numerator ...

$(x^2-6x+9)-[x^2+2xh+h^2 -6x-6h+9] = x^2-6x+9-x^2-2xh-h^2+6x+6h-9 = -2xh-h^2+6h = h(-2x-h+6)$

do not expand the denominator ... one of the (x-3)'s will divide out.
• Nov 27th 2010, 02:14 PM
wair
Yes I realized thank you I ended up with the answer you gave me earlier. so I'm pretty sure it is correct thank you.