Question

y=log√1-sin2x/1+sin2x then find dy/dx w.r.t.x​

Answer 1

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.