assume y is a differentiable function of x and dy/dx - sin(ay) + cos(bx)=xy

Answer 1

d/dx y^n = ny^(n-1)y'

d/dx x^n = nx^(n-1)

When differentiating 'implicitly,' you gotta pay attention to the y's

So, for you question, we have

d/dx [ sin(ay) + cos(bx) = xy ]

implies cos(ay)d/dx(ay) - sin(bx)d/dx(bx) = xd/dxy + yd/dxx

implies ay'cos(ay) - bsin(bx) = xy' + y

Then solve for y. I will leave the rest for you to do. I made a video, but it got caught in limbo. Not sure if it will pop up or not...

JJ