Given $y=4 x^{2}+3 x$, find $\frac{d y}{d t}$ when $x=-4$ and $\frac{d x}{d t}=5$
$\frac{\mathrm{dy}}{\mathrm{dt}}=\square($ Simplify your answer.)

Step 1 ：Given the function \(y = 4x^{2} + 3x\), we first need to find the derivative of \(y\) with respect to \(x\), denoted as \(\frac{dy}{dx}\).

Step 2 ：Using the power rule, the derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx} = 8x + 3\).

Step 3 ：We are given that \(\frac{dx}{dt} = 5\).

Step 4 ：By the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). Substituting the values we have, \(\frac{dy}{dt} = (8x + 3) \cdot 5 = 40x + 15\).

Step 5 ：We are asked to find \(\frac{dy}{dt}\) when \(x = -4\). Substituting \(x = -4\) into the equation, we get \(\frac{dy}{dt} = 40(-4) + 15 = -145\).

Step 6 ：So, the derivative of \(y\) with respect to \(t\) when \(x = -4\) and \(\frac{dx}{dt} = 5\) is \(\boxed{-145}\).