can anyone help me with this problem. I'm really stuck...?

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-15.jpg

Answer 1

Hint:

Derivative of
arcsin x
is
(1-x^2)^(-1/2)

Answer 2

f(x) -1/x = arcsin(x^2-4)/(x^2+4)
f(x) = 1/x +arcsin (x^2-4)/(x^2+4)
f(x) = 1/x +arcsin (1-8/(x^2+4))
You can take it from here.

Answer 3

sin[f(x)-1/x] = (x^2 - 4)/(x^2+4) ...(1)

Derivative (1) and get

[f'(x)+1/x^2]cos[f(x)-1/x] = [2x(x^2+4)-2x(x^2-4)]/(x^2+4)^2
[f'(x)+1/x^2]cos[f(x)-1/x] =16x/(x^2+4)^2 ...(2)

From cos(z) = Sqrt [1-sin^2 (z) ] where z = f(x)-1/x ...(3)

Substitute (1) in (3)
cos(z) = Sqrt [ 1 - {(x^2 - 4)/(x^2+4)}^2]
= Sqrt [ {(x^2+4)^2 - (x^2-4)^2} / (x^2+4)}^2]
= Sqrt [ 16x^2/(x^2+4)}^2]
cos(z) = 4x/(x^2+4) ... (4)

Substitute (4) in (2)

[f'(x)+1/x^2][4x/(x^2+4)] = 16x/(x^2+4)^2
[f'(x)+1/x^2] = 4/(x^2+4)
f'(x) = 4/(x^2+4) - 1/x^2
= (3x^2 - 4) / [(x^2)(x^2+4)]

The ans is 1.

Answer 4

f(x) = arcsin{(x^2-4)/X^2+4)] + 1/x
f'(x) = 1/sqrt(1-u^2)*du/dx - x^(-2)
= 1/sqrt[16x^2/(x^2+4)^2] * (16x/(x^2+4)^2 - x^(-2)
= (x^2+4)/4x * (16x/(x^2+4)^2 - x^(-2)
= 4/(x^2+4) - x^(-2)
= (4x^2-x^2 +4) / x^2(x^2+4)
= (3x^2 + 4) / x^2(x^2 + 4)