given ( y = 5x^{2}+x ), find ( \frac{dy}{dt} ) when ( x=-1 ) and ( \frac{dx}{dt}=4 ).\n( \frac{dy}{dt}=square…

given ( y = 5x^{2}+x ), find ( \frac{dy}{dt} ) when ( x=-1 ) and ( \frac{dx}{dt}=4 ).\n( \frac{dy}{dt}=square ) (simplify your answer.)

given ( y = 5x^{2}+x ), find ( \frac{dy}{dt} ) when ( x=-1 ) and ( \frac{dx}{dt}=4 ).\n( \frac{dy}{dt}=square ) (simplify your answer.)

Answer

Explanation:

Step1: Differentiate (y = 5x^{2}+x) with respect to (t)

Using the chain - rule (\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}). Differentiate (y) with respect to (x): (\frac{dy}{dx}=\frac{d}{dx}(5x^{2}+x)=10x + 1).

Step2: Substitute (x=-1) and (\frac{dx}{dt}=4)

First, find (\frac{dy}{dx}) at (x = - 1): (\frac{dy}{dx}\big|_{x=-1}=10(-1)+1=-10 + 1=-9). Then, use (\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}). Substitute (\frac{dy}{dx}=-9) and (\frac{dx}{dt}=4) into the formula. (\frac{dy}{dt}=(-9)\times4).

Answer:

(-36)