Help! dy/dx of inverse trig functions

[ugly_boy]

I must be missing something here. The derivatives of the inverse trig functions seemed pretty straightforward but none of my answers are even close to the ones in the book. Hopefully somebody can show me how to work through this one and I can try to do the rest once I see a good example.

Images after the cut.

Comments

[joshua_green]

Try expanding the quadratic in your denominator.

[korean_guy_01]

Upon FOILing the (2x+1)2 and then distributing the negative sign, I get the book's answer except with a 2 in front of the radical (2/4 = 1/2). That answer seems very close to me.

[joshua_green]

But there's a square root in the denominator, so you should get 2/√4 = 1.

[korean_guy_01]

Bad AlgebraI habit creeping back.

[alteredmana]

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

So you were right, it's just "simplification".