Question

given ( y = 7x^{2}+6x ), find ( \frac{dy}{dt} ) when ( x = - 1 ) and ( \frac{dx}{dt}=3 ).

( \frac{dy}{dt}=square ) (simplify your answer.)

Explanation:

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

Use the chain - rule \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\).
Differentiate \(y = 7x^{2}+6x\) with respect to \(x\): \(\frac{dy}{dx}=14x + 6\).

Step2: Substitute \(x=-1\) into \(\frac{dy}{dx}\)

When \(x = - 1\), \(\frac{dy}{dx}=14(-1)+6=-14 + 6=-8\).

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

Since \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\) and \(\frac{dx}{dt}=3\), \(\frac{dy}{dt}=(-8)\times3\).

Answer:

\(-24\)