If I know $ sinh^{-1}(x) $ = $ ln(x+ (1 + x^{2} ) ^ { \frac{1}{2} } ) $
I know a way to evaluate that derivative, it is : $\frac {1}{(1 + x^{2} )^{\frac{1}{2}} }$ but is it acceptable to take $ \frac{d}{dx} $ both sides above and work it out?
I have attempted it anyhow on RHS and here are the steps so far :
$ \frac{d}{dx} ln(x+ (1 + x^{2} ) ^ { \frac{1}{2} } ) = \frac { f'(x) } { f(x) } [ $ where $ f (x) = (x+ (1 + x^{2} ) ^ { \frac{1}{2} } ) $ ] $ = \frac { 1+ \frac{(x^{2} +1)^{\frac{-3}{2}}}{2} }{x+ (x^{2}+1)^{\frac{1}{2}}} $
and this is where I stopped.
Edit :
the reason was a simple mistake in differentation in the last step where it should be $ \frac {-1}{2} $ instead of $ \frac {-3}{2} $ which lead to problems with the simplification. Mark's answer is complete.
We have the following:
$$f(x)=sinh^{-1}(x)=log(x+(x^2+1)^{1/2})$$
So, taking the derivative of both sides yields:
$$\frac {dy}{dx}=\frac {1+\frac {x}{(x^2+1)^{1/2}}}{x+(1+x^2)^{1/2}}$$
This is simply via the chain rule, noting $\frac {d}{dx}log(x)=\frac 1x$, so $\frac {d}{dx}log(g(x))=\frac {g'(x)}{g(x)}$
So, combine the two fractions in the numerator into a single fraction:
$$\frac {dy}{dx}=\frac {(x^2+1)^{1/2}+x}{(x^2+1)^{1/2}}\cdot \frac {1}{(1+x^2)^{1/2}+x}=\frac {1}{(x^2+1)^{1/2}}$$