Find derivative of f(x) = 1/(3u^2 - 1)^2
f'(x) = (3u^2 - 1)^-2
= (-2)(3u^2 - 1)^-3(6u)
= -12u(3u^2 - 1)^-3
You probably mean,Originally Posted by becky
$\displaystyle y=\frac{1}{(3x^2-1)^2}$
Let,
$\displaystyle u=3x^2-1$
Then,
$\displaystyle y=\frac{1}{u^2}=u^{-2}$
Thus,
$\displaystyle \frac{dy}{du}=-2u^{-3}$
And,
$\displaystyle \frac{du}{dx}=6x$
Thus,
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
Which is,
$\displaystyle -2(6x)u^{-3}$
Substitute for u and simplify,
$\displaystyle -12x(3x^2-1)^{-3}$