Find dy/dx by implicit differentiation. cos(xy) = 1 + sin(y)

Answer 1

Have you tried to differentiate either side of the equation?

Answer 2

Let's look at the left hand side of the equation first.

d/dx [ cos(xy) ] = -sin(xy) * d/dx(xy) (chain rule)

= -sin(xy) * [y + x * dy/dx ] (product rule)

= -sin(xy) * y - sin(xy) * x * dy/dx (segregating terms with & without dy/dx)

derivative of right hand side is just sin(y) * dy/dx

Therefore (equating derivatives of both sides):

-sin(xy) * y - sin(xy) * x * dy/dx = sin(y) * dy/dx

So -sin(xy) * y = [sin(xy) * x + sin(y)] * dy/dx

and finally dy/dx = - [sin(xy) * y] / [sin(xy) * x + sin(y)]