Question

Question #af0a8

Answer 1

`-1`

Explanation:

Making `z=x+y` we have

`sinz=logz rArr z=e^(sinz)`

This equation has a solution as can be depicted in the attached figure (red - `y=e^(sinz)`, blue `y=x`) so calling this solution as `z_0` we have

`x+y = z_0` and now

`dx + dy = 0 rArr dy/dx = -1`