Asked by C.S.
deriv.
y=x(sinx)(cosx)
y=x(sinx)(cosx)
Answers
Answered by
bobpursley
y=uvw
y'=vwu' + uvw'+uwv'
It is just algebra, and probably some trig identities.
y'=vwu' + uvw'+uwv'
It is just algebra, and probably some trig identities.
Answered by
Steve
use the product rule.
hard way:
(1)[(sinx)(cosx)] + {x(cosx)](cosx) + [x(sinx)](-sinx)
= sinx cosx + xcos^2 x - xsin^2 x
= 1/2 sin 2x + xcos 2x
easier way:
y = 1/2 x sin 2x
y' = 1/2 (sin 2x) + 1/2 x (2cos 2x)
= 1/2 sin 2x + x cos 2x
hard way:
(1)[(sinx)(cosx)] + {x(cosx)](cosx) + [x(sinx)](-sinx)
= sinx cosx + xcos^2 x - xsin^2 x
= 1/2 sin 2x + xcos 2x
easier way:
y = 1/2 x sin 2x
y' = 1/2 (sin 2x) + 1/2 x (2cos 2x)
= 1/2 sin 2x + x cos 2x
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