Second derivatives Find d2ydx2 x+y=sin⁡y

Question

Second derivatives Find d2ydx2  x+y=sin⁡y

Asma Vang · Accepted Answer

Step 1
First derivative is the rate of change in y w.r.t x, and
Second derivative is rate of change in

\frac{d y}{d x}

w.r.t x.
Derivative of the function decides whether the function is increasing or decreasing. If the first derivative is positive, function is increasing in that particular interval. If it is negative, function is decreasing for that particular interval.
Step 2
Given function is

x + y = \sin y

differentiating both sides w.r.t x , we get

1 + \frac{d y}{d x} = \cos y \cdot \frac{d y}{d x}

\frac{d y}{d x} = \frac{1}{\cos y - 1}

Differentiating again w.r.t x, we get

\frac{d^{2} y}{{d x}^{2}} = - \sin y \cdot \frac{d y}{d x} \cdot \frac{d y}{d x} + \cos y \frac{d^{2} y}{{d x}^{2}}

\frac{d^{2} y}{{d x}^{2}} (1 - \cos y) = - \sin y \cdot {(\frac{d y}{d x})}^{2}

\frac{d^{2} y}{{d x}^{2}} = \frac{- \sin y \cdot {(\frac{d y}{d x})}^{2}}{1 - \cos y}

\frac{d^{2} y}{{d x}^{2}} = \frac{- \sin y \cdot {(\frac{1}{1 - \cos y})}^{2}}{1 - \cos y}

= - \frac{\sin y}{{(1 - \cos y)}^{3}}

Second derivatives Find \frac{d^{2}y}{dx^{2}} x+y=\sin y

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