# Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0?

Jan 4, 2018

$x = 1$, when $y = 0$

$y ' ' \left(1\right) = - 3$

#### Explanation:

Given:

ln(x) + y = ln(x^2) − y^2

Add ${y}^{2} - \ln \left(x\right)$ to both sides:

y^2 + y = ln(x^2) − ln(x)

Use the fact that subtraction of logarithms is the as division within the argument:

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

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

${y}^{2} + y - \ln \left(x\right) = 0 \text{ [1]}$

Here is a graph of equation [1]:

We can use the quadratic formula to obtain two equations of y in terms of x

$y = - \frac{1}{2} + \frac{\sqrt{1 + 4 \ln \left(x\right)}}{2} \text{ [1a]}$

and

$y = - \frac{1}{2} - \frac{\sqrt{1 + 4 \ln \left(x\right)}}{2} \text{ [1b]}$

Here are the graphs of equations [1a] and [1b]:

Please observe that equation [1a] contains the point of interest, $\left(1 , 0\right)$, therefore, we shall only be differentiating this equation.

$y ' = \frac{1}{x \sqrt{4 \ln \left(x\right) + 1}}$

y''= -(4ln(x)+3)/(x^2(4ln(x)+1)^(3/2)

Evaluate when $y = 0$ and $x = 1$:

$y ' ' \left(1\right) = - 3$