Find the derivative of function with respect to x

[togo]

Find the derivative of function with respect to x

y = sin^-1 (x-1/x+1)

Steps I took:
= sin(x-1/x+1)^-1
u = sin(x-1/x+1)
u' = cos u * u'
u = x-1/x+1
u' = t'b - tb'/b^2

t = x-1
b = x+1
t' = 1
b' = 1
b^2 = (x+1)(x+1) = x^2+2x+1
u' = (1)(x+1) - (x-1)(1)/x^2+2x+1
u' = x+1-x-1 / x^2+2x+1
u' = 0 !

how can it be zero? something has gone wrong here.

The answer to the question of 1/(x^(1/2) (x+1))

Thank you

 

[mfb]

There are missing brackets: x-1/x+1 is $x-\frac{1}{x}+1$. I think you mean (x-1)/(x+1), or $\frac{x-1}{x+1}$.

u' = ((1)(x+1) - (x-1)(1))/(x^2+2x+1)
u' = (x+1-x-1) / (x^2+2x+1)

(I added brackets here)
This step is wrong, check the signs in the numerator.

 

[SteamKing]

Is the initial function sin or arcsin?

If it is arcsin, I don't think you have the correct derivative.

 

[togo]

how would you tell the difference? (sin/arcsin)

 

[eumyang]

how would you tell the difference? (sin/arcsin)

sin^-1 x is the same as arcsin x. I'm assuming that your original function is
$y = sin^{-1}\left( \frac{x-1}{x+1} \right)$
so that looks like arcsin. Of course, the derivative of arcsin is not cosine.