Question

How do you implicitly differentiate `-1=sinxy`?

Answer 1

`y'=-y/x`

Explanation:

`-1=sin(xy)`

Differentiate implicitly with respect to `x`:

`D(-1)=D(sinxy)`

Use the chain rule on `d(sin(xy))/(dx)rArr`

`0=cos(xy)xx(xy'+y)`

`:.0=cos(xy)xxxy'+cos(xy)xxy`

`:.cancel(cos(xy))xxxy'=cancel(-cos(xy))xxy`

`:.xy'=-y`

`:.y'=-y/x`