1. ## derivative of arctan

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

$\displaystyle \arctan(\sqrt{1+x}/\sqrt{1-x})$

2. Originally Posted by IrrationalPI3
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

$\displaystyle \arctan(\sqrt{1+x}/\sqrt{1-x})$

Recall that

$\displaystyle \frac{d}{dx}\left[\arctan{u}\right]=\frac{u'}{1+u^2}$

You have $\displaystyle u=\frac{\sqrt{1+x}}{\sqrt{1-x}}$

So plug-n-play.

3. Originally Posted by IrrationalPI3
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 $\displaystyle \frac{1}{{2\sqrt {1 - {x^2}} }}.$

4. wow that was fast thanks people, i will try the method offered and check my answer against the one given. Thanks again

I guess high school is gonna be boring now that i can do this stuff already lol

5. With your attitude, you're gonna go places.

Stick around here at MHF, we need people like you.