y=log√1-sin2x/1+sin2x then find dy/dx w.r.t.x
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Step-by-step explanation:
Let u = 1-sin2x & v = 1+sin2x
and let y = u½v-½ and
thus u'=-2cos2x, v'=2cos2x
Then by product rule,
y' = (d/dx)u½•v-½ + u½•(d/dx)v-½
....
=½(1-sin2x)-½(-2cos2x)(1+sin2x)-½
-½(1-sin2x)½(1+sin2x)-(3/2)(2cos2x)
=(-cos2x)(1+sin2x)-(3/2)(1-sin2x)-½ ...factored, & times [(1+sin2x) + (1-sin2x)]
which I simplify to -2cos2x over
(1+sin2x)(3/2)(1-sin2x)½
and that's as far as I wish to go! Check it very carefully because it was a nightmare to key in...and of course this is not the completed job.
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