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