D/dx(sin^4(cot^-1(1-x/1+x)^1/2)))?

1 Answer
Nov 25, 2017

1/2(1+x).

Explanation:

Given that, y=sin^4(cot^-1sqrt{(1-x)/(1+x)}), we need dy/dx.

We substitute x=cos2theta," so that, "-1 lt x lt 1.

Note that, in order to make sqrt((1-x)/(1+x)) meaningful, we must have

-1 lt x lt 1, which justifies our substitution : x=cos2theta.

:. 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.!