h(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]
41) Find the derivative.
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.
1 answer