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