If sin²y + cosxy = k, find at x = 1,

We are given with an equation sin²y + cos(xy) = k , we have to find at x = 1, y = by using the given equation, so by differentiating the equation on both sides with respect to x, we get,

2siny cosy – sin(xy)[(1)y + x] = 0

[2siny cosy – xsin(xy)] = ysin(xy)

By putting the value of point in the derivative, which is x = 1, y = ,

_{(x = 1,y = π/4)} =

_{(x = 1,y = π/4)} =