how do we do this derivative problem?

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-14.jpg

Answer 1

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

(x^2 - 4) / (x^2 + 4)
= (x^2 - 4 + 4 - 4) / (x^2 + 4)
=(x^2 + 4 - 8) / (x^2 + 4)
= 1 - 8(x^2 + 4)^-1

y - (1/x) = arcsin [1 - 8(x^2 + 4)^-1]

y = arcsin [1 - 8(x^2 + 4)^-1] + 1/x

derivative of arcsin(u) is 1/ sqrt(1- u^2)
In this case, u = 1 - 8(x^2 + 4)^-1

So let's take this piece by piece:
(1 - 8(x^2 + 4)^-1) ^2 = 1 -16(x^2 + 4)^-1 + 64(x^2 + 4)^-2

1 - (1 -16(x^2 + 4)^-1 + 64(x^2 + 4)^-2) = 16(x^2 + 4)^-1 - 64(x^2 + 4)^-2
= [16(x^2 + 4) - 64] / (x^2 + 4)^2
= 16[x^2 + 4 - 4] / (x^2 + 4)^2
= (16x^2) / (x^2 + 4)^2

sqrt[(16x^2) / (x^2 + 4)^2] = 4x / (x^2 + 4)

1/ [4x / (x^2 + 4)] = (x^2 + 4) / 4x

Now apply the chain rule:
derivative of 1 - 8(x^2 + 4)^-1
= 8 (x^2 + 4) ^-2 * 2x
= 16x (x^2 + 4)^ -2

[(x^2 + 4) / 4x] * [16x (x^2 + 4)^ -2]
= 4 / (x^2 + 4)

Wrapping it up:
derivative of 1/x
= -1/ (x^2)

So,
y = arcsin [1 - 8(x^2 + 4)^-1] + 1/x
y' = [4/ (x^2 + 4)] - [1/ (x^2)]

Simplify:
=[(4)(x^2) - (x^2 + 4)] / [(x^2)(x^2 + 4)]
=[ 4x^2 - x^2 - 4] / [(x^2)(x^2 + 4)]
=[ 3x^2 - 4] / [(x^2)(x^2 + 4)]

That was one of the most difficult derivatives I've ever taken, and I would guess there is an easier way to do it. I did try doing implicit differentiation, but that was even more difficult.

Answer 2

Great work by Kyle. Here is the cheat answer. If it's a multiple choice and you're not required to show working, then here is a shortcut involving the definition of differentation.

f(x) = 1/x + arcsin[(x^2 - 4)/(x^2 + 4)]
f(5) = 1.0098
f(5.1) = 1.01942
f(4.9) = 0.99983
So f'(5) ~= (1.01942 - 0.99983) / 0.2 = +0.09795

Plug in x = 5 into the answer choices and #5 works.