CalculusDerivatives
posted by Charles .
How is 6#sin(#x1)/cos^3(#x1) the derivative of f(x)= 3sec^2(#x1)?
#= Pi (I coulnd't find the symbol...sorry.)

recall that d(secu)/dx
= tanu secu du/dx
so if f(x) = 3sec^2(pix1)
= 3(sec(pix1))^2 then using the chain rule
f'(x) = 6(sec(pix1)*tan(pix1)sec(pix1)*pi
= 6pi(sin(pix1)/cos(pix1)*sec^2(pix1)
= 6pi(sin(pix1))/cos(pix1)*1/cos^2(pix1)
= 6pi(sinpix1)/cos^3(pix1)