Question

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

\(-38\)