# Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0? Note that the domain for x is that x > 0. There will only be one point on the curve with y = 0

Jul 25, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{4 {y}^{2} + 4 y + 3}{{x}^{2} {\left(1 + 2 y\right)}^{3}}$

For $x = 1$ we have $y \left(1\right) = 0$ and $y ' ' \left(1\right) = - 3$

#### Explanation:

Given the implicit function:

$\ln x + y = \ln \left({x}^{2}\right) - {y}^{2}$

defined for $x > 0$, note that based on the properties of logarithms:

$\ln \left({x}^{2}\right) = 2 \ln x$

so that the equation becomes:

${y}^{2} + y = \ln x$

and then for $x = 1$ we have $y = 0$.

Differentiate both sides of the equation:

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

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

and again:

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

substituting:

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

we have:

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

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{{x}^{2} {\left(1 + 2 y\right)}^{3}} \left(2 + {\left(1 + 2 y\right)}^{2}\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{{x}^{2} {\left(1 + 2 y\right)}^{3}} \left(2 + 1 + 4 y + 4 {y}^{2}\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{4 {y}^{2} + 4 y + 3}{{x}^{2} {\left(1 + 2 y\right)}^{3}}$

For $x = 1$:

${\left[\frac{{d}^{2} y}{\mathrm{dx}} ^ 2\right]}_{x = 1} = - 3$

Jul 25, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \pm 3$

#### Explanation:

$\log x + y = \log {x}^{2} - {y}^{2} \Rightarrow \log \left(x {e}^{y}\right) = \log \left({x}^{2} {e}^{- {y}^{2}}\right)$

Assuming $x > 0$ we have

$x {e}^{y} = {x}^{2} {e}^{- {y}^{2}}$ or

${e}^{y + {y}^{2}} = x$ or

${y}^{2} + y - \log x = 0$ now solving for $y$

$y = \frac{1}{2} \left(- 1 \pm \sqrt{1 + 4 \log x}\right)$ which is real for $1 + 4 \log x \ge 0$ or

$0 < x$ and now

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \pm \frac{3 + 4 L o g x}{{x}^{2} {\left(1 + 4 L o g x\right)}^{\frac{3}{2}}}$

Now for ${e}^{y + {y}^{2}} = x$ we have $y = 0 \Rightarrow x = 1$ and then

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \pm 3$