Derivative of $y-2\sin(y)=x$ defines: $y=y(x)$ I need to find the derivative of $y'$ and $y''$ given that:
$$y-2\sin(y)=x$$
defines:
$$y=y(x)$$
Thank you!
 A: HINT:
Differentiating both sides of the given equation with respect to $x$ reveals
$$y'(x)-2\cos(y)y'(x)=1 \tag 1$$
SPOLIER ALERT:  Scroll over the highlighted area to reveal the solution.

Solving $(1)$ for $y'(x)$ yields $$y'(x)=\frac{1}{1-2\cos(y)} \tag 2$$Now, differentiating $(2)$ with respect to $x$ yields $$\begin{align}y''(x)&=-\frac{2\sin(y)y'(x)}{(1-2\cos(y))^2}\\\\&=-\frac{2\sin(y)}{(1-2\cos(y))^3}\\\\&=\frac{x-y}{(1-2\cos(y))^3}\end{align}$$ 

A: $$\\ { y }^{ \prime  }-2\cos { \left( y \right) { y }^{ \prime  } } =1\\ { y }^{ \prime  }=\frac { 1 }{ 1-2\cos { \left( y \right)  }  } \\ { y }^{ \prime \prime  }-2\left( -\sin { \left( y \right) { \left( { y }^{ { \prime  } } \right)  }^{ 2 }+\cos { \left( y \right) { y }^{ \prime \prime  } }  }  \right) =0\\ $$
A: HINT:
Differentiate both sides of equation, primes w.r.t. $x $:
$$y'-2\cos y \,y'=1  $$
$$  y^{\prime} = \frac{1}{1-2 \cos y} \tag{1} $$
Constant is 1, so we can in the quotient / product rule setting differentiate:
$$ \frac{y^{\prime}}{1-2 \cos y}=   -\frac{y^{\prime\prime}}{2 \sin y} $$
Plug in for $y^{\prime}$from (1)
$$   \frac{1}{(1-2 \cos y)^2} =   -\frac{y^{\prime\prime}}{2 \sin y} \tag{2}$$
which is same result as the others.
