Question

How do you differentiate `y=x^3/y+3xy`?

Answer 1

`y' = (3(x^2y + y^3))/ (y^2-3xy^2 + x^3)`

Explanation:

`(y=x^3/y+3xy)'`

`\implies y' = (x^3/y)' + (3xy)'`

from quotient rule
`(x^3/y)' = (3x^2 y - x^3 y')/y^2`

from product rule
`(3xy)' = 3y + 3 xy'`

`\implies y' = (3x^2 y - x^3 y')/y^2 + 3y + 3 xy'`

`\implies y'(1-3x + x^3/y^2) = 3((x^2 )/y + y)`

`\implies y'(y^2-3xy^2 + x^3) = 3(x^2y + y^3)`

`\implies y' = (3(x^2y + y^3))/ (y^2-3xy^2 + x^3)`