Question

Question #b46eb

Answer 1

`1.dy/dx=(y-2x)/(2y-x)`
`2. dy/dx =(sin y/x^2)/([cosy/x-1/y^2])`

Explanation:

`1. d/dx[x^2+y^2=xy+3]`
`2x+2y dy/dx=xdy/dx +y`
`2y dy/dx -xdy/dx = y-2x`
`dy/dx (2y-x)=y-2x`
`dy/dx=(y-2x)/(2y-x)`

`2. siny/x+1/y = 1`
`d/dx[x^-1siny+y^-1 = 1]`
`1/xcosy dy/dx -siny/x^2 -1/y^2dy/dx=0`
`1/xcosy dy/dx -1/y^2dy/dx = sin y/x^2`
`dy/dx [cosy/x-1/y^2]=sin y/x^2`
`dy/dx =(sin y/x^2)/([cosy/x-1/y^2])`