Find y′y'y′ if y=(x2+1)3xy = \frac{(x^2+1)^3}{x}y=x(x2+1)3.
(x2+1)2(5x2−1)x2\frac{(x^2+1)^2(5x^2-1)}{x^2}x2(x2+1)2(5x2−1)
3(x2+1)2(2x)3(x^2+1)^2(2x)3(x2+1)2(2x)
6x(x2+1)2⋅x−(x2+1)3x2\frac{6x(x^2+1)^2 \cdot x - (x^2+1)^3}{x^2}x26x(x2+1)2⋅x−(x2+1)3
Both (a) and (c) are equivalent