• Jan 23rd 2007, 03:13 AM
My brain Gave up on this problem....
Find the derivative:
\$\displaystyle D_x(3x^3 + x^{-3})(x+3)(x^2-5)\$
• Jan 23rd 2007, 03:38 AM
CaptainBlack
Quote:

My brain Gave up on this problem....
Find the derivative:
\$\displaystyle D_x(3x^3 + x^{-3})(x+3)(x^2-5)\$

Use the product rule:

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

......\$\displaystyle =(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
\$\displaystyle 18x^5+45x^4-60x^3-120x^2-3x^{-2}-5x^{-3}\$
• Jan 23rd 2007, 06:02 AM
Soroban

Quote:

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

There is an extended product rule . . .

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

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

. . Get the pattern?

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