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April 19, 2015

April 19, 2015

Posted by **Steff** on Saturday, June 11, 2011 at 5:02pm.

h(x)= (3x^2 + 1)^3 / (x^2 - 1)^4

This is where I am at the moment....

h'(x)= (x^2-1)^4 (3)(3x^2+1)^2(6x) - (3x^2+1)^3(4)(x^2-1)^3(2x)

This is what I am trying to get....(not final answer)

=2x(x^2-1)^3(3x^2+1)^2[(9)(x^2-1) - (4)(3x^2+1)}

Am really struggling with this. I have the answer in the back of the book but am not getting it right. have the solutions manual also. The chapter is on general power rule and chain rule.

- derivative -
**Damon**, Saturday, June 11, 2011 at 5:16pmh(x)= (3x^2 + 1)^3 / (x^2 - 1)^4

rewrite as

h(x)= (3x^2 + 1)^3 * (x^2 - 1)^-4

first*derivative second + second*derivative first

h' = (3x^2 + 1)^3*(-4)(x^2 - 1)^-5 (2x)

+(x^2 - 1)^-4 *(3)(3x^2 + 1)^2 (6x)

=(3x^2 + 1)^2(x^2 - 1)^-4 [-8x(3x^2 + 1)(x^2 - 1)^-1 +12x]

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