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Let $x$ and $y$ be functions of $t$ with $y = 2x^3 + 4x^2 + 3$ . If $\frac{dx}{dt} = \frac{1}{3}$ , what is $\frac{dy}{dt}$ when $x = 4$ ? $\newline$ Write an exact, simplified answer.

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Write and solve direct variation equations

Full solution

Q. Let

x

and

y

be functions of

t

with

y = 2x^3 + 4x^2 + 3

. If

\frac{dx}{dt} = \frac{1}{3}

, what is

\frac{dy}{dt}

when

x = 4

?

\newline

Write an exact, simplified answer.

Identify Function: Identify the function and differentiate it with respect to $x$ . Given $y = 2x^3 + 4x^2 + 3$ , differentiate each term with respect to $x$ . $\frac{dy}{dx} = 6x^2 + 8x$
Apply Chain Rule: Apply the chain rule to find $\frac{dy}{dt}$ . Using $\frac{dy}{dt} = \left(\frac{dy}{dx}\right) \cdot \left(\frac{dx}{dt}\right)$ , substitute $\frac{dx}{dt} = \frac{1}{3}$ . $\frac{dy}{dt} = \left(6x^2 + 8x\right) \cdot \left(\frac{1}{3}\right)$
Substitute $x = 4$ : Substitute $x = 4$ into the expression for $\frac{dy}{dt}$ .
$\frac{dy}{dt} = (6*(4)^2 + 8*4) * (\frac{1}{3})$
$\frac{dy}{dt} = (6*16 + 32) * (\frac{1}{3})$
$\frac{dy}{dt} = (96 + 32) * (\frac{1}{3})$
$\frac{dy}{dt} = 128 * (\frac{1}{3})$
$\frac{dy}{dt} = 42.67$

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