derivative

**becky** · September 3rd 2006, 07:29 PM

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

**ThePerfectHacker** · September 3rd 2006, 07:32 PM

Originally Posted by becky

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,
$y=\frac{1}{(3x^2-1)^2}$
Let,
$u=3x^2-1$
Then,
$y=\frac{1}{u^2}=u^{-2}$
Thus,
$\frac{dy}{du}=-2u^{-3}$
And,
$\frac{du}{dx}=6x$
Thus,
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
Which is,
$-2(6x)u^{-3}$
Substitute for u and simplify,
$-12x(3x^2-1)^{-3}$

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