# Derive arctangent

• Aug 1st 2007, 11:29 AM
wizzler
Derive arctangent
I want to derive this equation:

$
y*arctan(x-2) + x*arctan(\frac{y-1}{20}) = 0
$

should i first use the product formulae on the terms on each side of the addition sign? Later should I use the sum formulae on that?
• Aug 1st 2007, 11:33 AM
Krizalid
You have to use implicit differentiation, the product rule and

$(\arctan{u})'=\frac1{1+u^2}\cdot{u'}$
• Aug 1st 2007, 11:39 AM
wizzler
Quote:

Originally Posted by Krizalid
You have to use implicit differentiation, the product rule and

$(\arctan{u})'=\frac1{1+u^2}\cdot{u'}$

yea and what happens to the y and the x before the arctan term?
• Aug 1st 2007, 11:47 AM
tukeywilliams
Quote:

Originally Posted by wizzler
I want to derive this equation:

$
y*arctan(x-2) + x*arctan(\frac{y-1}{20}) = 0
$

should i first use the product formulae on the terms on each side of the addition sign? Later should I use the sum formulae on that?

You use the product rule.

So $\frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\left(\frac{y-1}{20} \right)^2 \right)} + \arctan \left(\frac{y-1}{20} \right)$
• Aug 1st 2007, 11:12 PM
earboth
Quote:

Originally Posted by tukeywilliams
You use the product rule.

So $\frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\frac{y-1}{20} \right)^2} + \arctan \left(\frac{y-1}{20} \right)$

Hello,

I don't want to pick at you ... but I can't see you you around here so I'm going to remove the small typo in your result:

$\frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\left(\frac{y-1}{20} \right)^2\right)} + \arctan \left(\frac{y-1}{20} \right)$