[Cambridge/9231/2007]
The positive variables $x$ and $y$ are related by $y = x^2 + 2\ln(xy).$ Find the values of $\frac{\text{d}y}{\text{d}x}$ and $\frac{\text{d}^2y}{\text{d}x^2}$ when both $x$ and $y$ are equal to 1.

1. Expand the logarithmic term and differentiate the equation implicitly with respect to $x$.

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

Differentiating with respect to $x$:
$\frac{\text{d}y}{\text{d}x} = 2x + \frac{2}{x} + \frac{2}{y}\frac{\text{d}y}{\text{d}x}$

2. Substitute the given point $(1, 1)$ into the derivative equation to find $\frac{\text{d}y}{\text{d}x}$.

Substitute $x = 1$ and $y = 1$:
$\frac{\text{d}y}{\text{d}x} = 2(1) + \frac{2}{1} + \frac{2}{1}\frac{\text{d}y}{\text{d}x}$
$\frac{\text{d}y}{\text{d}x} = 4 + 2\frac{\text{d}y}{\text{d}x}$
$\frac{\text{d}y}{\text{d}x} - 2\frac{\text{d}y}{\text{d}x} = 4$
$-\frac{\text{d}y}{\text{d}x} = 4$
$\frac{\text{d}y}{\text{d}x} = -4$

3. Differentiate the first derivative equation implicitly with respect to $x$ to find an expression involving $\frac{\text{d}^2y}{\text{d}x^2}$.

From $\frac{\text{d}y}{\text{d}x} = 2x + 2x^{-1} + 2y^{-1}\frac{\text{d}y}{\text{d}x}$, differentiate both sides with respect to $x$ using the product rule on the last term:
$\frac{\text{d}^2y}{\text{d}x^2} = 2 - 2x^{-2} + 2 \left[ y^{-1}\frac{\text{d}^2y}{\text{d}x^2} - y^{-2}\left(\frac{\text{d}y}{\text{d}x}\right)^2 \right]$

4. Substitute $x = 1$, $y = 1$, and $\frac{\text{d}y}{\text{d}x} = -4$ into the second derivative equation to find $\frac{\text{d}^2y}{\text{d}x^2}$.

$\frac{\text{d}^2y}{\text{d}x^2} = 2 - 2(1)^{-2} + 2\left[ (1)^{-1}\frac{\text{d}^2y}{\text{d}x^2} - (1)^{-2}(-4)^2 \right]$
$\frac{\text{d}^2y}{\text{d}x^2} = 0 + 2\left[ \frac{\text{d}^2y}{\text{d}x^2} - 16 \right]$
$\frac{\text{d}^2y}{\text{d}x^2} = 2\frac{\text{d}^2y}{\text{d}x^2} - 32$
$\frac{\text{d}^2y}{\text{d}x^2} - 2\frac{\text{d}^2y}{\text{d}x^2} = -32$
$-\frac{\text{d}^2y}{\text{d}x^2} = -32$
$\frac{\text{d}^2y}{\text{d}x^2} = 32$