Question

Find dy/ddx for xy² - sin(x+2y)= 2x?

Answer 1

`dy/dx=(2-y^2+cos(x+2y))/(2xy +2)`

Explanation:

Starting from the equation

`xy^2-sin(x+2y)=2x`

We differentiate both sides of the equation, using `d/dx`

`d/dx(xy^2-sin(x+2y))=d/dx(2x)`
`d/dx(xy^2)-d/dx(sin(x+2y))=d/dx(2x)`
`x d/dx(y^2) + y^2(1)-cos(x+2y)d/dx(x+2y)=2`
`x 2ydy/dx + y^2-cos(x+2y)d/dx(x)+d/dx(2y)=2`
`x 2ydy/dx + y^2-cos(x+2y)+2dy/dx=2`
`2xydy/dx +2dy/dx=2-y^2+cos(x+2y)`
`dy/dx(2xy +2)=2-y^2+cos(x+2y)`
`dy/dx=(2-y^2+cos(x+2y))/(2xy +2)`