Question

given ( y = 2x^{2}+3x ), find ( \frac{dy}{dt} ) when ( x=-3 ) and ( \frac{dx}{dt}=4 ).
( \frac{dy}{dt}=square ) (simplify your answer.)

Explanation:

Step1: Differentiate \(y\) with respect to \(t\)

Using the chain - rule \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\).
Differentiate \(y = 2x^{2}+3x\) with respect to \(x\): \(\frac{dy}{dx}=4x + 3\).

Step2: Substitute \(x=-3\) into \(\frac{dy}{dx}\)

When \(x = - 3\), \(\frac{dy}{dx}=4(-3)+3=-12 + 3=-9\).

Step3: Calculate \(\frac{dy}{dt}\)

Since \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\) and \(\frac{dx}{dt}=4\), then \(\frac{dy}{dt}=(-9)\times4\).

Answer:

\(-36\)