hey everyone just doing some calc for fun again :P and i ran across another question i wasnt sure how to do any help would be greatly appreciated
find the derivative of
$\displaystyle \arctan(\sqrt{1+x}/\sqrt{1-x})
$
$\displaystyle \arctan \sqrt {\frac{{1 + x}}{{1 - x}}}.$
$\displaystyle \frac{d}
{{dx}}\arctan \sqrt {\frac{{1 + x}}
{{1 - x}}} = {\left[ {1 + {{\left( {\sqrt {\frac{{1 + x}}
{{1 - x}}} } \right)}^2}} \right]^{ - 1}} \cdot \frac{d}
{{dx}}\sqrt {\frac{{1 + x}}
{{1 - x}}} =$
$\displaystyle = \frac{1}
{2}{\left( {1 + \frac{{1 + x}}
{{1 - x}}} \right)^{ - 1}} \cdot \left[ {{{\left( {\frac{{1 + x}}
{{1 - x}}} \right)}^{ - 1/2}} \cdot \frac{d}
{{dx}}\left( {\frac{{1 + x}}
{{1 - x}}} \right)} \right] =$
$\displaystyle = \frac{{1 - x}}
{4} \cdot \frac{{\sqrt {1 - x} }}
{{\sqrt {1 + x} }} \cdot \frac{2}
{{{{\left( {1 - x} \right)}^2}}} = \ldots$
Now simplify
Your answer must be $\displaystyle \frac{1}{{2\sqrt {1 - {x^2}} }}.$