y=(sinx/4)^4 +(cosx/4)^4
那么
y'=4(sinx/4)^3 *(sinx/4)' + 4(cosx/4)^3 *(cosx/4)'
显然
(sinx/4)'=1/4* cosx/4,(cosx/4)'= -1/4 *sinx/4
所以
y'=4(sinx/4)^3 *1/4* cosx/4 -4(cosx/4)^3 *1/4 *sinx/4
=(sinx/4)^3 *cosx/4 -(cosx/4)^3 *sinx/4
=sinx/4 *cosx/4 *[(sinx/4)^2 - (cosx/4)^2]
= 0.5sinx/2 * (-cosx/2)
= -1/4 *sinx
即y' = -1/4 *sinx好棒！不必客气的啊～