- #1
PhysicsinCalifornia
- 58
- 0
Please Help Me !
Hello,
I REALLY need help deriving a function
if [tex]f(x) = \frac{-x-2}{(x-2)^3}[/tex]
what would its derivative be?
Here's what I did so far::
Using the quotient rule,
[tex]f'(x) = \frac{(x-2)^3(-1) - (-x-2)(3(x-2)^2(1))}{((x-2)^3)^2)}[/tex]
[tex]= \frac{-(x-2)^3 + (x+2)(3(x^2-4x+4))}{(x-2)^6}[/tex]
[tex]= \frac{-(x^2 -4x+4)(x-2) + (x+2)(3x^2 -12x +12)}{(x-2)^6}[/tex]
[tex]= \frac{-(x^3 -4x^2 + 4x -2x^2 +8x -8) + (3x^3 -12x^2 + 12x +6x^2 -24x +24)}{(x-2)^6}[/tex]
[tex]= \frac{-x^3 + 4x^2 -2x + 2x^2 -8x + 8 + 3x^3 - 12x^2 + 12x + 6x^2 -24x +24}{(x-2)^6}[/tex]
[tex]f'(x) = \frac{2x^3 - 24x + 32}{(x-2)^6}[/tex]
Is this the derivative, and if so, how can I find the critical numbers? (When the derivative equals 0 or DNE(does not exist))
Hello,
I REALLY need help deriving a function
if [tex]f(x) = \frac{-x-2}{(x-2)^3}[/tex]
what would its derivative be?
Here's what I did so far::
Using the quotient rule,
[tex]f'(x) = \frac{(x-2)^3(-1) - (-x-2)(3(x-2)^2(1))}{((x-2)^3)^2)}[/tex]
[tex]= \frac{-(x-2)^3 + (x+2)(3(x^2-4x+4))}{(x-2)^6}[/tex]
[tex]= \frac{-(x^2 -4x+4)(x-2) + (x+2)(3x^2 -12x +12)}{(x-2)^6}[/tex]
[tex]= \frac{-(x^3 -4x^2 + 4x -2x^2 +8x -8) + (3x^3 -12x^2 + 12x +6x^2 -24x +24)}{(x-2)^6}[/tex]
[tex]= \frac{-x^3 + 4x^2 -2x + 2x^2 -8x + 8 + 3x^3 - 12x^2 + 12x + 6x^2 -24x +24}{(x-2)^6}[/tex]
[tex]f'(x) = \frac{2x^3 - 24x + 32}{(x-2)^6}[/tex]
Is this the derivative, and if so, how can I find the critical numbers? (When the derivative equals 0 or DNE(does not exist))
Last edited: