Consider the function β(π₯) = 1 + sin(ππ₯) Find ββ²(1). β(π₯) = 1 + sin(ππ₯) β'(π₯) = 0 + πβcos(ππ₯) β'(1) = πβcos(πβ1) β'(1) = πβcos(π) β'(1) = πβ(-1) β'(1) = -π What's all that about g? g is not involved with h or h' Edwin