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})

$

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- Oct 1st 2009, 04:53 PMIrrationalPI3derivative 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})

$ - Oct 1st 2009, 04:58 PMVonNemo19
- Oct 1st 2009, 05:19 PMDeMath
$\displaystyle \arctan \sqrt {\frac{{1 + x}}{{1 - x}}}.$

$\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 $\displaystyle \frac{1}{{2\sqrt {1 - {x^2}} }}.$ - Oct 1st 2009, 05:27 PMIrrationalPI3
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, 05:29 PMVonNemo19
With your attitude, you're gonna go places.

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