# Thread: Problem with Diff'ing a rational expression

1. ## 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

2. 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

3. Hello, BClary!

You seem to be doing fine . . .

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 . . .

4. ## Thanks to all, off doing second now

Thanks for all your help, lets see where I end up at

thanks

5. ## 2nd derivative should be

after a little algrebra (haha)

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

Right?

6. Originally Posted by BClary
after a little algrebra (haha)

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

Right?
Yup!

-Dan

7. ## Thanks so much

TY Ty Ty Ty ty TY