Let x and y be functions of t with y = x^3 + 2x + 1. If dx/dt = -1, what is dy/dt when x = 4? Write an exact, simplified answer. ______

Apply Chain Rule: To find dy/dt, use the chain rule on y = x^3 + 2x + 1. Differentiate y with respect to x to get dy/dx. dy/dx = 3x^2 + 2.Substitute x = 4: Substitute x = 4 into dy/dx to find the value at x = 4. dy/dx = 3(4)^2 + 2 = 3*16 + 2 = 48 + 2 = 50.Use Chain Rule with dx/dt: Now, use the given dx/dt = -1 to find dy/dt using the chain rule: dy/dt = dy/dx * dx/dt. dy/dt = 50 * (-1) = -50.

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

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Q. Let

x

and

y

be functions of

t

with

y = x^3 + 2x + 1

. If

\frac{dx}{dt} = -1

, what is

\frac{dy}{dt}

when

x = 4

\newline

Write an exact, simplified answer.

Apply Chain Rule: To find $\frac{dy}{dt}$ , use the chain rule on $y = x^3 + 2x + 1$ . Differentiate $y$ with respect to $x$ to get $\frac{dy}{dx}$ . $\newline$ $\frac{dy}{dx} = 3x^2 + 2$ .
Substitute $x = 4$ : Substitute $x = 4$ into $\frac{dy}{dx}$ to find the value at $x = 4$ . $\frac{dy}{dx} = 3(4)^2 + 2 = 3\times16 + 2 = 48 + 2 = 50$ .
Use Chain Rule with $\frac{dx}{dt}$ : Now, use the given $\frac{dx}{dt} = -1$ to find $\frac{dy}{dt}$ using the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ . $\frac{dy}{dt} = 50 \cdot (-1) = -50$ .