Question

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

\(-36\)