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

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

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

Answer

Explanation:

Step1: Differentiate (y = 2x^{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}(2x^{2}+x)=4x + 1).

Step2: Substitute (x=-5) into (\frac{dy}{dx})

When (x = - 5), (\frac{dy}{dx}=4(-5)+1=-20 + 1=-19).

Step3: Use the chain - rule formula (\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt})

Given (\frac{dx}{dt}=2), then (\frac{dy}{dt}=(-19)\times2).

Answer:

(-38)