Question

hi, "Differentiate: y=[v^3-2v(v)^(1/2)]/v
ty!

Answer 1

let me see if i get this, \[dy/dv [(v^3-2v(v^.5)/(v)]\] if so first simplify the top to v^3-2v^3/2 then turn the whole equation into (v^3-2v^3/2)(v^-1) then use the chain rule to get (3v^2-3v^.5)(v^-1)+(v^3-2v^3/2)(v^-2)(-1) then use algebra to simplify

Answer 2

\[y=[v^3-2v(v)^0.5]/v\]
it it your equation?

Answer 3

this is the answer: \[Y'= [2v^3-v^1.5]/v^2\]