If $\log y = \log(\sin x) - x^2$, then $\dfrac{d^2y}{dx^2} + 4x\dfrac{dy}{dx} + 4x^2y =$

Given $\log y = \log(\sin x) - x^2$

Differentiating with respect to $x$:

$\dfrac{1}{y} \dfrac{dy}{dx} = \cot x - 2x$

$\dfrac{dy}{dx} = y(\cot x - 2x)$

Differentiating again with respect to $x$:

$\dfrac{d^2y}{dx^2} = \dfrac{dy}{dx} \cot x - y \csc^2 x - 2x \dfrac{dy}{dx} - 2y$

The given expression is $\dfrac{d^2y}{dx^2} + 4x\dfrac{dy}{dx} + 4x^2y$. Substituting $\dfrac{d^2y}{dx^2}$ into the expression:

$\dfrac{d^2y}{dx^2} + 4x\dfrac{dy}{dx} + 4x^2y = \left( \dfrac{dy}{dx} \cot x - y \csc^2 x - 2x \dfrac{dy}{dx} - 2y \right) + 4x\dfrac{dy}{dx} + 4x^2y$

$= \dfrac{dy}{dx} \cot x - y \csc^2 x + 2x \dfrac{dy}{dx} - 2y + 4x^2y$

$= \dfrac{dy}{dx} (\cot x + 2x) - y \csc^2 x - 2y + 4x^2y$

Substituting $\dfrac{dy}{dx} = y(\cot x - 2x)$:

$= y(\cot x - 2x)(\cot x + 2x) - y \csc^2 x - 2y + 4x^2y$

$= y(\cot^2 x - 4x^2) - y \csc^2 x - 2y + 4x^2y$

$= y \cot^2 x - 4x^2y - y \csc^2 x - 2y + 4x^2y$

$= y(\cot^2 x - \csc^2 x) - 2y$

Using the identity $\csc^2 x - \cot^2 x = 1$, we get $\cot^2 x - \csc^2 x = -1$:

$= y(-1) - 2y = -3y$

Answer: $-3y$