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September 16, 2014

September 16, 2014

Posted by **Charles** on Sunday, September 28, 2008 at 10:02pm.

#= Pi (I coulnd't find the symbol...sorry.)

- Calculus-Derivatives -
**Reiny**, Sunday, September 28, 2008 at 11:35pmrecall that d(secu)/dx

= tanu secu du/dx

so if f(x) = 3sec^2(pix-1)

= 3(sec(pix-1))^2 then using the chain rule

f'(x) = 6(sec(pix-1)*tan(pix-1)sec(pix-1)*pi

= 6pi(sin(pix-1)/cos(pix-1)*sec^2(pix-1)

= 6pi(sin(pix-1))/cos(pix-1)*1/cos^2(pix-1)

= 6pi(sinpix-1)/cos^3(pix-1)

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