# Math Help - derivative

1. ## derivative

trying to differentiate:
$1-x^2 / (x^2+1)^2$

using the quotient rule,

$u = (1-x^2)$
$du/dx = -2x$

$v = (x^2+1)^2)$
$dv/dx = 4x(1+x^2)$

$[u * dv/dx ] - [v *du/dx] / v^2$

I've been given an answer to complete working out to:
$2x(x^2-3)/(x^2+1)^3$

I've managed to work it out up to
$2x(x^2-1)/(x^2+1)^3$

but have no idea where the 3 in the the provided answer came from

2. Originally Posted by JohnnyB
trying to differentiate:
$1-x^2 / (x^2+1)^2$

using the quotient rule,

$u = (1-x^2)$
$du/dx = -2x$

$v = (x^2+1)^2)$
$dv/dx = 4x(1+x^2)$

$[u * dv/dx ] - [v *du/dx] / v^2$

I've been given an answer to complete working out to:
$2x(x^2-3)/(x^2+1)^3$

I've managed to work it out up to
$2x(x^2-1)/(x^2+1)^3$

but have no idea where the 3 in the the provided answer came from
$\frac{d}{dx} \left[\frac{1-x^2}{(x^2+1)^2}\right]$

$\frac{(x^2+1)^2 \cdot (-2x) - (1-x^2) \cdot 4x(x^2+1)}{(x^2+1)^4}$

$\frac{(x^2+1) \cdot (-2x) - (1-x^2) \cdot 4x}{(x^2+1)^3}
$

$\frac{-2x^3 - 2x - 4x + 4x^3}{(x^2+1)^3}$

$\frac{2x^3 - 6x}{(x^2+1)^3}$

$\frac{2x(x^2-3)}{(x^2+1)^3}$

3. uuughhhhhh, so easy.

made a typo in my working out.