2007-09-21 18:14:27 · 3 answers · asked by Rob 1 in Science & Mathematics ➔ Mathematics
sin^40xcos40x
= (1/2) sin^80x
d/dx [ (1/2) sin^80x ]
= (1/2) * (80) sin^(79x) * cosx
=40 sin^79x cosx
2007-09-21 18:22:14 · answer #1 · answered by Madhukar 7 · 1⤊ 1⤋
Use d(uv)= udv + vdu
let
u = sin^40 x
du = 40sin^39 x (cosx)
v= cos40x
dv= -40sinxcosx
=sin^40x(-40sinxcosx) + cos40x(40sin^39 x cos x)
=-40 sin^41xcosx + 40sin^39 x cos^2 40x
2007-09-22 01:36:57 · answer #2 · answered by Ojo 2 · 0⤊ 1⤋
Use the product rule and the chain rule.
d[(sin^40 x)(cos 40x)] / dx
= 40[(sin^39 x)(cos x)(cos 40x)] + (sin^40 x)[-40(sin 40x)]
= 40[(sin^39 x)(cos x)(cos 40x)] - 40[(sin^40 x)(sin 40x)]
= 40(sin^39 x) [(cos x)(cos 40x) - (sin x)(sin 40x)]
Notice the angle addition formula.
= 40(sin^39 x)(cos 41x)
2007-09-22 02:55:45 · answer #3 · answered by Northstar 7 · 1⤊ 0⤋