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given ( y = 2x^{2}+5x ), find ( \frac{dy}{dt} ) when ( x=-5 ) and ( \fr…

Question

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

Explanation:

Step1: Differentiate \(y\) with respect to \(t\)

Using the chain - rule \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\).
First, find \(\frac{dy}{dx}\) for \(y = 2x^{2}+5x\).
\(\frac{dy}{dx}=\frac{d}{dx}(2x^{2}+5x)=4x + 5\).

Step2: Substitute \(x=-5\) and \(\frac{dx}{dt}=2\)

Substitute \(x = - 5\) into \(\frac{dy}{dx}\): \(\frac{dy}{dx}\big|_{x=-5}=4(-5)+5=-20 + 5=-15\).
Then, since \(\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\), and \(\frac{dx}{dt}=2\).
\(\frac{dy}{dt}=(-15)\times2\).

Answer:

\(-30\)