Thread: 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

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

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

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

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

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

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

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

-Dan

Hello, BClary!

You seem to be doing fine . . .

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

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

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

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

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

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


Now try the second derivative . . .

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

thanks

after a little algrebra (haha)

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

Right?

Originally Posted by BClary

after a little algrebra (haha)

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

Right?

Yup!

-Dan

TY Ty Ty Ty ty TY