# PLease Help Me!

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• Jan 23rd 2007, 03:13 AM
^_^Engineer_Adam^_^
PLease Help Me!
My brain Gave up on this problem....
Find the derivative:
$D_x(3x^3 + x^{-3})(x+3)(x^2-5)$
Thanks ahead :)
• Jan 23rd 2007, 03:38 AM
CaptainBlack
Quote:

Originally Posted by ^_^Engineer_Adam^_^
My brain Gave up on this problem....
Find the derivative:
$D_x(3x^3 + x^{-3})(x+3)(x^2-5)$
Thanks ahead :)

Use the product rule:

$D_x(3x^3 + x^{-3})(x+3)(x^2-5)=$ $
D_x[3x^3+x^{-3}]\ (x+3) (x^2-5) + (3x^3+x^{-3}) D_x[(x+3)(x^2-5)]$

...... $=(9x^2-3x^{-4})(x+3)(x^2-5) +(3x^3+x^{-3}) [(x^2-5)+(x+3)(2x)]$

I will leave simplifying this further to you.

RonL
• Jan 23rd 2007, 04:47 AM
^_^Engineer_Adam^_^
My answer is:
$18x^5+45x^4-60x^3-120x^2-3x^{-2}-5x^{-3}$
• Jan 23rd 2007, 06:02 AM
Soroban
Hello, ^_^Engineer_Adam^_^!

Quote:

Find the derivative: . $y \:=\:(3x^3 + x^{-3})(x+3)(x^2-5)$

There is an extended product rule . . .

If $y \:=\:f\!gh$, then: . $y' \;=\;f'gh + f\!g'h + f\!gh'$

If $y \:=\:f\!ghk$, then: . $y' \;=\;f'ghk + f\!g'hk + f\!gh'k + f\!ghk'$

. . Get the pattern?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you're going to mutlply-out your final answer,
. . you might as well multiply first . . .

We have: . $y \;=\;3x^6 + 9x^5 - 15x^4 - 45x^3 + 1 - 3x^{-1} - 5x^{-2} - 15x^{-3}$

Then: . $y' \;=\;18x^5 + 45x^4 - 60x^3 - 135x^2 + 3x^{-2} + 10x^{-3} + 45x^{-4}$