# derivative of arctan

• Oct 1st 2009, 05:53 PM
IrrationalPI3
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

$\arctan(\sqrt{1+x}/\sqrt{1-x})
$
• Oct 1st 2009, 05:58 PM
VonNemo19
Quote:

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

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

Recall that

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

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

So plug-n-play.
• Oct 1st 2009, 06:19 PM
DeMath
Quote:

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

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

$\arctan \sqrt {\frac{{1 + x}}{{1 - x}}}.$

$\frac{d}
{{dx}}\arctan \sqrt {\frac{{1 + x}}
{{1 - x}}} = {\left[ {1 + {{\left( {\sqrt {\frac{{1 + x}}
{{1 - x}}} } \right)}^2}} \right]^{ - 1}} \cdot \frac{d}
{{dx}}\sqrt {\frac{{1 + x}}
{{1 - x}}} =$

$= \frac{1}
{2}{\left( {1 + \frac{{1 + x}}
{{1 - x}}} \right)^{ - 1}} \cdot \left[ {{{\left( {\frac{{1 + x}}
{{1 - x}}} \right)}^{ - 1/2}} \cdot \frac{d}
{{dx}}\left( {\frac{{1 + x}}
{{1 - x}}} \right)} \right] =$

$= \frac{{1 - x}}
{4} \cdot \frac{{\sqrt {1 - x} }}
{{\sqrt {1 + x} }} \cdot \frac{2}
{{{{\left( {1 - x} \right)}^2}}} = \ldots$

Now simplify

Your answer must be $\frac{1}{{2\sqrt {1 - {x^2}} }}.$
• Oct 1st 2009, 06:27 PM
IrrationalPI3
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
• Oct 1st 2009, 06:29 PM
VonNemo19
With your attitude, you're gonna go places.

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