What is the derivative of (x^2-1)^3/(x^2+1)(x2−1)3x2+1?

1 Answer
ali ergin
May 22, 2016

(d y)/(d x)=-2x*(x^2-1)^3/(x^2+1)^2+6x(x^2-1)^2/(x^2+1dydx=−2x⋅(x2−1)3(x2+1)2+6x(x2−1)2x2+1

Explanation:

d/(d x)(x^2-1)^3/(x^2+1)=?ddx(x2−1)3x2+1=?

d/(d x) (x^2-1)^3/(x^2+1) = (3(x^2-1)^2*2x*(x^2+1)-2x(x^2-1)^3)/(x^2+1)^2ddx(x2−1)3x2+1=3(x2−1)2⋅2x⋅(x2+1)−2x(x2−1)3(x2+1)2

d/(d x)(x^2-1)^3/(x^2+1)=-2x*(x^2-1)^3/(x^2+1)^2+6x(x^2-1)^2/(x^2+1)ddx(x2−1)3x2+1=−2x⋅(x2−1)3(x2+1)2+6x(x2−1)2x2+1

Related questions
See all questions in Quotient Rule
Impact of this question
1493 views around the world
You can reuse this answer
Creative Commons License