How do you differentiate g(y) =(y +2 y^3)(2y^2 -3y^3) using the product rule?

1 Answer

y'=y(-36y^4+20y^3-12y^2+6y)

Explanation:

(y+2y^3)=u
(2y^2-3y^3)=v
Using product rule
(dy)/dx=V(du)/dx+U(dv)/dx
(du)/dx=(1+6y^2);(dv)/dx=(4y-9y^2)
Hence
(dy)/dx=(2y^2-3y^3)(1+6y^2)+(y+2y^3)(4y-9y^2)
=>2y^2+12y^4-3y^3-18y^5+4y^2-9y^3+8y^4-18y^5
Collecting like terms
=>-36y^5+20y^4-12y^3+6y^2
=>y(-36y^4+20y^3-12y^2+6y)
Hope it helps