Find the second derivative

G

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Find the second Derivative

f(x)=1/x^2 -1

I am not getting the right answer and not sure where I am messing up.

f'(x)=(x^2-1)(1)-1(2x)/ (x^2-1)^2
=x^2-1-2x/(x^2-1)^2

f''=(x^2-1)(-2)(-2x)


I am having trouble with the second derivative.

Thank for any help you may give.

Amy
 

ChaoticLlama

Junior Member
Joined
Dec 11, 2004
Messages
199
I'm assuming your function is:

f(x)=1/(x² - 1)

when the numerator is a constant, use product rule

f(x) = (x² - 1)^-1

now differentiate..

f'(x) = -1(x² - 1)^-2 *(2x)

f'(x) = -2x(x² - 1)^-2

now do it again for the second derivative.This time you should do quotient rule normally.

f'(x) = -2x / (x² - 1)²

f''(x) = {(x² - 1)² * (-2) - (-2x) * 2 * (x² - 1) * (2x)} / (x² - 2)^4

f''(x) = {-2(x² - 1)² + 8x²(x² - 1)} / (x² - 2)^4

f''(x) = {-2(x² - 1)² + 8x²(x² - 1)} / (x² - 2)^4

Now just factor the numerator and simplify.
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
Another way is u-substitution. It's mostly associated with integration but it works for differentiation too.

\(\displaystyle y=\frac{1}{x^{2}-1}\)

Let u=\(\displaystyle x^{2}-1\); \(\displaystyle du=2xdx\)

\(\displaystyle \frac{du}{u}=\frac{-1}{u^{2}}\)

Sub back in:

dy=\(\displaystyle \frac{-2x}{(x^{2}-1)^{2}}dx\)

\(\displaystyle \frac{dy}{dx}=\frac{-2x}{(x^{2}-1)^{2}}\)

You could use the product rule now for the 2nd derivative:

\(\displaystyle (-2x)(x^{2}-1)^{-2}dx\)

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

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