How is 6#sin(#x-1)/cos^3(#x-1) the derivative of f(x)= 3sec^2(#x-1)?

Question

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

Answer 1

recall 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)