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September 14, 2014

September 14, 2014

Posted by **Kira** on Friday, May 18, 2012 at 3:54pm.

Can someone help me with this? Thanks.

- Calculus AB -
**bobpursley**, Friday, May 18, 2012 at 4:11pmI am not certain if the 20pi x^2 is in the denominator or not of the first term. If it is really this,

f(x)=[1/(x(27,000,000/pi + 6,750,000)] + 20PI x^2

then work it in two parts, the first, and second term.

f(x)=1/kx + 20 pi x^2

f'= -1/kx^2 + 40PI x

where k= 27E6/PI +6.75E6

- Calculus AB -
**Kira**, Friday, May 18, 2012 at 4:16pmHere I'll write it again but distribute the 1/x so you can see it better.

f(x) = (27,000,000/pix)+(6,750,000/x)+20pix^2

Does that help?

- Calculus AB -
**Kira**, Friday, May 18, 2012 at 4:17pmThe 1/x was only factored out of the first two terms, not out of the

20pi(x^2)

- Calculus AB -
**Kira**, Friday, May 18, 2012 at 4:31pmBut I get the general idea. I just write 1/x as x^-1 and differentiate that. Makes sense, thanks!

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