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I need to find the derivative of $y'$ and $y''$ given that:

$$y-2\sin(y)=x$$

defines: $$y=y(x)$$

Thank you!

mvw
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BAM
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  • Implicit differentiation w.r.t. $x$ returns an equation from which you can solve out $y'(x)$. –  Jun 13 '16 at 17:19

3 Answers3

2

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 $$\begin{align}y''(x)&=-\frac{2\sin(y)y'(x)}{(1-2\cos(y))^2}\\\\&=-\frac{2\sin(y)}{(1-2\cos(y))^3}\\\\&=\frac{x-y}{(1-2\cos(y))^3}\end{align}$$

Mark Viola
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$$\\ { y }^{ \prime }-2\cos { \left( y \right) { y }^{ \prime } } =1\\ { y }^{ \prime }=\frac { 1 }{ 1-2\cos { \left( y \right) } } \\ { y }^{ \prime \prime }-2\left( -\sin { \left( y \right) { \left( { y }^{ { \prime } } \right) }^{ 2 }+\cos { \left( y \right) { y }^{ \prime \prime } } } \right) =0\\ $$

haqnatural
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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.

Narasimham
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