Question

Question #338b6

Answer 1

`\frac{dy}{dx} = \frac{-y^3cos(x) + sin(x - 2y)}{3y^2sin(x) + 2sin(x - 2y)}`

Explanation:

`y^3sin(x) + cos(x - 2y) = 5`

Take derivative of both sides:

`3y^2 \frac{dy}{dx}sin(x) + y^3cos(x) - sin(x - 2y)(1 - 2\frac{dy}{dx}) = 0`

Solve for dy/dx:

`3y^2 \frac{dy}{dx}sin(x) + y^3cos(x) - sin(x - 2y) + 2sin(x - 2y)\frac{dy}{dx} = 0`

`3y^2\frac{dy}{dx}sin(x)+ 2sin(x-2y)\frac{dy}{dx} = -y^3cos(x)+ sin(x-2y)`

#\frac{dy}{dx}(3y^2sin(x) + 2sin(x - 2y)) = -y^3cos(x) + sin(x - 2y)#

`\frac{dy}{dx} = \frac{-y^3cos(x) + sin(x - 2y)}{3y^2sin(x) + 2sin(x - 2y)}`