Question

Differentiate: y = (v^3 - 2v[sqrt[v]])/v

Answer 1

wow confusing

Explanation:

Answer 2

`y =(v^3 - 2v√(v))/(v)=\frac{v^3 - 2v^{(3)/(2)}}{v} \\ \\y ='(\frac{v^3 - 2v^{(3)/(2)}}{v})'= \frac{(v^3 - 2v^{(3)/(2)})'v-(v^3 - 2v^{(3)/(2)})v^' }{v^2} = ((3v^2-3 √(v))v-(v^3 - 2v √(v) ) )/(v^2) = \\ \\=((3v^2-3 √(v))v-v(v^2 - 2 √(v) ) )/(v^2) =(v(3v^2-3 √(v)-v^2 +2 √(v) ) )/(v^2) =(2v^2- √(v) )/(v) =(2v√(v)-1 )/( √(v) )`