Question

Question #43877

Answer 1

`dy/dx=(cos(x+y)-cos(y))/(-xsin(y)-cos(x+y))`

Explanation:

Differentiate both sides with respect to `x`. This means that every time we differentiate a term containing `y`, we should end up with an instance of `dy/dx`.

`d/dx(xcos(y))=d/dx(sin(x+y))`

`cos(y)d/dx(x)+xd/dxcos(y)=cos(x+y)d/dx(x+y)`

`-xsin(y)dy/dx+cos(y)=(1+dy/dx)cos(x+y)`

Solve for `dy/dx:`

`-xsin(y)dy/dx+cos(y)=cos(x+y)+dy/dxcos(x+y)` (Multiply out on the right side)

`-xsin(y)dy/dx-dy/dxcos(x+y)=cos(x+y)-cos(y)` (Isolate all terms containing `dy/dx` on the left side, move all other terms to the right)

`dy/dx(-xsin(y)-cos(x+y))=cos(x+y)-cos(y)` (Factor out `dy/dx`)

`dy/dx=(cos(x+y)-cos(y))/(-xsin(y)-cos(x+y))`
(Divide both sides by `-xsin(y)-cos(x+y)`)