Question

How do you implicitly differentiate `-1=(x-y)sinx+y`?

Answer 1

`dy/dx = (sinx+xcosx-ycosx)/ (sinx -1)`

Explanation:

`-1=(x-y)sinx+y`

`-1=xsinx-ysinx+y`

`0=sinx+xcosx-ycosx-sinx dy/dx+dy/dx`

`sinx dy/dx-dy/dx=sinx+xcosx-ycosx`

`dy/dx (sinx-1)=sinx+xcosx-ycosx`

`dy/dx = (sinx+xcosx-ycosx)/ (sinx -1)`