Given that, y=sin^4(cot^-1sqrt{(1-x)/(1+x)}),y=sin4(cot−1√1−x1+x), we need dy/dx.dydx.
We substitute x=cos2theta," so that, "-1 lt x lt 1.x=cos2θ, so that, −1<x<1.
Note that, in order to make sqrt((1-x)/(1+x))√1−x1+x meaningful, we must have
-1 lt x lt 1,−1<x<1, which justifies our substitution : x=cos2theta.x=cos2θ.
:. y=sin^4(cot^-1sqrt{(1-x)/(1+x)}),
=sin^4(cot^-1sqrt{(1-cos2theta)/(1+cos2theta)}),
=sin^4(cot^-1sqrt{(2sin^2theta)/(2cos^2theta)}),
=sin^4(cot^-1(tantheta)),
=sin^4(cot^-1{cot(pi/2-theta)}),
=sin^4(pi/2-theta),
={sin(pi/2-theta)}^4,
=cos^4theta,
=(cos^2theta)^2,
={(1+cos2theta)/2}^2.
rArr y=1/4(1+x)^2........................................[because, cos2theta=x].
:. dy/dx=1/4*d/dx(1+x)^2,
=1/4*2(1+x)*d/dx(1+x)..............[because," the Chain Rule],"
rArr dy/dx=1/2(1+x).
Enjoy Maths.!
