# How do you find the second derivative of y^2 + x + sin y = 9?

Aug 4, 2015

The second derivative is $y ' ' = \frac{\sin \left(y\right) - 2}{2 y + \cos \left(y\right)} ^ 3$

#### Explanation:

Implicit differentiation gives us

$\frac{d}{\mathrm{dx}} \left({y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x\right) + \frac{d}{\mathrm{dx}} \left(\sin \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(9\right)$
$2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 + \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$2 y \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + \cos \left(y\right)\right) = - 1$. Therefore,

$y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{2 y + \cos \left(y\right)}$. This implies that $2 y + \cos \left(y\right) \ne 0$.

Taking the second implicit derivative gives us
$\frac{d}{\mathrm{dx}} \left[\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + \cos \left(y\right)\right) = - 1\right]$

By application of the Product Rule, we have
$\frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(2 y + \cos \left(y\right)\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 \frac{\mathrm{dy}}{\mathrm{dx}} - \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 \left(2 y + \cos \left(y\right)\right) + 2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} - \sin \left(y\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 \left(2 y + \cos \left(y\right)\right) = \sin \left(y\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} - 2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\sin \left(y\right) {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} - 2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}}{\left(2 y + \cos \left(y\right)\right)}$ or more succinctly

$y ' ' = {\left(y '\right)}^{2} \frac{\sin \left(y\right) - 2}{2 y + \cos \left(y\right)}$
$y ' ' = {\left(\frac{- 1}{2 y + \cos \left(y\right)}\right)}^{2} \frac{\sin \left(y\right) - 2}{2 y + \cos \left(y\right)}$
$y ' ' = \frac{1}{2 y + \cos \left(y\right)} ^ 2 \frac{\sin \left(y\right) - 2}{2 y + \cos \left(y\right)}$
$y ' ' = \frac{\sin \left(y\right) - 2}{2 y + \cos \left(y\right)} ^ 3$

Aug 4, 2015

I get: $y ' ' = \frac{- 2 + \sin y}{2 y + \cos y} ^ 3$

#### Explanation:

$x + {y}^{2} + \sin y = 9$

$\frac{d}{\mathrm{dx}} \left(x + {y}^{2} + \sin y\right) = \frac{d}{\mathrm{dx}} \left(9\right)$

$1 + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{2 y + \cos y} = - {\left(2 y + \cos y\right)}^{-} 1$

$\frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1 {\left(2 y + \cos y\right)}^{-} 2 \cdot \left[\frac{d}{\mathrm{dx}} \left(2 y + \cos y\right)\right]$

$= {\left(2 y + \cos y\right)}^{-} 2 \left[2 \frac{\mathrm{dy}}{\mathrm{dx}} - \sin y \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$= {\left(2 y + \cos y\right)}^{-} 2 \left[2 - \sin y\right] \frac{\mathrm{dy}}{\mathrm{dx}}$

$= {\left(2 y + \cos y\right)}^{-} 2 \left[2 - \sin y\right] \left[- {\left(2 y + \cos y\right)}^{-} 1\right]$

$= - \left(2 - \sin y\right) {\left(2 y + \cos y\right)}^{-} 3$

$= \frac{- 2 + \sin y}{2 y + \cos y} ^ 3$