# Problem with Diff'ing a rational expression

• Nov 18th 2007, 07:17 AM
BClary
Problem with Diff'ing a rational expression
Keep getting all twisted up in the algebra in finding the first and second derivative of the following

x(third)-4x(squared)/(x-2)(squared)

F’(x)
(x(third)-4x(squared))’(x-2)(squared)-(x(third)-4x(squared))((x-2)(squared))’/(x-2)(fourth)

than

(3x(squared)-8x)(x-2)(squared)-(x(third)-4x(squared)(2)(x-2)(1)/(x-2)(fourth)

than

(x-2)[(3x(squared)-8x)(x-2)-2x(third)+8x(squared)]/(x-2)(fourth)

now stuck

any help on further factoring would be appreciated
• Nov 18th 2007, 07:45 AM
topsquark
Quote:

Originally Posted by BClary
Keep getting all twisted up in the algebra in finding the first and second derivative of the following

x(third)-4x(squared)/(x-2)(squared)

F’(x)
(x(third)-4x(squared))’(x-2)(squared)-(x(third)-4x(squared))((x-2)(squared))’/(x-2)(fourth)

than

(3x(squared)-8x)(x-2)(squared)-(x(third)-4x(squared)(2)(x-2)(1)/(x-2)(fourth)

than

(x-2)[(3x(squared)-8x)(x-2)-2x(third)+8x(squared)]/(x-2)(fourth)

now stuck

any help on further factoring would be appreciated

$F(x) = \frac{x^3 - 4x^2}{(x - 2)^2}$

So
$F^{\prime}(x) = \frac{(3x^2 - 8x)(x - 2)^2 - (x^3 - 4x^2)(2(x - 2))}{(x - 2)^4}$

$F^{\prime}(x) = \frac{(x - 2)[(3x^2 - 8x)(x - 2) - 2(x^3 - 4x^2)]}{(x - 2)^4}$

This is about where you got to. Now divide through by the common x - 2:
$F^{\prime}(x) = \frac{(3x^2 - 8x)(x - 2) - 2(x^3 - 4x^2)}{(x - 2)^3}$

Expand:
$F^{\prime}(x) = \frac{3x^3 - 8x^2 - 6x^2 + 16x - 2x^3 + 8x^2)}{(x - 2)^3}$

And simplify:
$F^{\prime}(x) = \frac{x^3 - 6x + 16}{(x - 2)^3}$

-Dan
• Nov 18th 2007, 08:05 AM
Soroban
Hello, BClary!

You seem to be doing fine . . .

Quote:

Find the first and second derivatives: . $F(x) \;=\;\frac{x^3-4x^2}{(x-2)^2}$

Quotient Rule: . $F'(x) \;=\;\frac{(x-2)^2\cdot(3x^2-8x) - (x^3-4x^2)\cdot2(x-2)}{(x-2)^4}$

. . . . . . . . . . . $F'(x) \;=\;\frac{x(x-2)^2(3x-8) - 2x^2(x-4)(x-2)} {(x-2)^4}$

Factor/reduce: . $F'(x) \;=\;\frac{x(x-2)\cdot\left[(x-2)(3x-8) - 2x(x-4)\right]}{(x-2)^4} \;=$ . $\frac{x\cdot\left[(x-2)(3x-8) - 2x(x-4)\right]}{(x-2)^3}$

Simplify: . . . . $F'(x) \;=\;\frac{x\left[3x^2 - 8x - 6x + 16 - 2x^2 + 8x\right]}{(x-2)^3}$

Therefore: . . . $\boxed{F'(x) \;=\;\frac{x\left(x^2-6x+16\right)}{(x-2)^3}}$

Now try the second derivative . . .

• Nov 18th 2007, 08:09 AM
BClary
Thanks to all, off doing second now
Thanks for all your help, lets see where I end up at

thanks
• Nov 18th 2007, 08:57 AM
BClary
2nd derivative should be
after a little algrebra (haha)

(-8x-32)/((x-2)(fourth))

Right?
• Nov 18th 2007, 08:59 AM
topsquark
Quote:

Originally Posted by BClary
after a little algrebra (haha)

(-8x-32)/((x-2)(fourth))

Right?

Yup! :)

-Dan
• Nov 18th 2007, 09:00 AM
BClary
Thanks so much
TY Ty Ty Ty ty TY