Question

# 5 pls !!
find dy/dx by implicit differentiation

Answer 1

`(dy)/(dx)` =
`(y-3x^2)/(2y-x)`

Explanation:

using the product rule to differentiate - xy then

3x² - (x
`(dy)/(dx)` + y(1) ) + 2y
`(dy)/(dx)` = 0

3x² - x
`(dy)/(dx)` - y + 2y
`(dy)/(dx)` = 0

3x² +
`(dy)/(dx)` (2y - x) - y = 0 (subtract 3x² - y from both sides )

`(dy)/(dx)` (2y - x) = y - 3x² ← divide both sides by (2y - x)

`(dy)/(dx)` =
`(y-3x^2)/(2y-x)`

Answer 2

Explanation:

`5. {x}^(3) - xy + {y}^(2) = 4`

`(dy)/(dx) ( {x}^(3) - xy + {y}^(2) ) = (dy)/(dx) (4)`

`3 {x}^(2) - x(1) (dy)/(dx) + 1(y) + 2y (dy)/(dx)`

Combine the dy/dx.

`(dy)/(dx) ( - x + 2y) + y + 3 {x}^(2)`

`(dy)/(dx) ( - x + 2y) = - 3 {x}^(2) - y`

`(dy)/(dx) = \frac{ - 3 {x}^(2) - y}{ - x + 2y}`

`\frac{3 {x}^(2) + y }{x - 2y}`