# Thread: derivative - using quotient rule

1. ## derivative - using quotient rule

how do I get the derivative of:

-4x
---------
(x^2-1)^2

12x^2+4
--------------
(x^2-1)^3

using the quotient rule, it seems simple enough but I cant seem to cancel out properly. Thanks!

2. Just in case a picture helps...

... where

... is the chain rule, wrapped inside...

... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

Similar to the quotient rule, but no automatic squaring of the denominator.
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3. Originally Posted by calculus0
how do I get the derivative of:

-4x
---------
(x^2-1)^2

12x^2+4
--------------
(x^2-1)^3

using the quotient rule, it seems simple enough but I cant seem to cancel out properly. Thanks!
$\displaystyle y = \frac{-4x}{(x^2 - 1)^2}$

$\displaystyle \frac{dy}{dx} = \frac{(x^2 - 1)^2\,\frac{d}{dx}(-4x) - (-4x)\,\frac{d}{dx}[(x^2 - 1)^2]}{[(x^2 - 1)^2]^2}$

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

$\displaystyle = \frac{-4(x^2 - 1)^2 + 16x^2(x^2 - 1)}{(x^2 - 1)^4}$

$\displaystyle = \frac{(x^2 - 1)[-4(x^2 - 1) + 16x^2]}{(x^2 - 1)^4}$

$\displaystyle = \frac{-4x^2 + 4 + 16x^2}{(x^2 - 1)^3}$

$\displaystyle = \frac{12x^2 + 4}{(x^2 - 1)^3}$

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