Find the range by finding the domain of the inverse y= (x+1) / (x^2 +1)?

I agree with Saurav, but I am putting steps for you to understand properly. Kindly learn it and do not keep asking here blankly. By Product rule for derivatives d/dx(uv) = u.

Dv/dx + v. Du/dx; where you and v are functions of x. In this case, you = cos(2x) and v = sin(1/2x) dy/dx cos(2x).

D/dxsin(1/2x) + sin(1/2x). D/dxcos(2x) By chain rule for derivatives cos(2x). Cos(1/2x).

D/dx(1/2x) + sin(1/2x). -sin(2x). D/dx(2x) cos(2x).

Cos(1/2x).(1/2)d/dx(x^-1) + sin(1/2x). -sin(2x).2 cos(2x). Cos(1/2x).(1/2)-(x^-2) + sin(1/2x).

-sin(2x).2 - (1/2).(x^-2). Cos(2x). Cos(1/2x) - 2.

Sin(2x). Sin(1/2x) - (1/2x^2). Cos(2x).

Cos(1/2x) - 2. Sin(2x). Sin(1/2x).

Y' = -0.5sin(2x)sin(1/x) + cos(2x).(-1/x^2)cos(1/x) just use product rule and next time before asking first go to wolfram and check out there!

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.