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

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

  • Calculus-Derivatives -

    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)

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