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\( 1 \leftarrow \quad \begin{array}{l}\text { Given the equation } 4 x^{3}+x^{2} y-2 y^{3}=0 \text {, evaluate } \frac{d y}{d x} \text {. Assume that the equation implicitly defines } y \text { as a differentiable } 1 \\ \text { function of } x \text {. } \\ \text { If } F(x, y)=4 x^{3}+x^{2} y-2 y^{3} \text {, then } F x=\square .\end{array} \)

Ask by Barker Martin. in the United States
Mar 17,2025

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\( F_x = 12x^2 + 2xy \).

Solución

We are given the function \[ F(x,y)=4x^3+x^2y-2y^3. \] We need to compute the partial derivative with respect to \( x \), treating \( y \) as a constant. 1. Differentiate \( 4x^3 \) with respect to \( x \): \[ \frac{\partial}{\partial x}(4x^3)=12x^2. \] 2. Differentiate \( x^2y \) with respect to \( x \): Since \( y \) is a constant, \[ \frac{\partial}{\partial x}(x^2y)=2xy. \] 3. Differentiate \(-2y^3\) with respect to \( x \): This term does not contain \( x \) so its derivative is 0. Thus, combining the results, we have \[ F_x=12x^2+2xy. \]

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To find \(\frac{dy}{dx}\) from the given equation \(4x^3 + x^2y - 2y^3 = 0\), we can use implicit differentiation. First, we differentiate both sides of the equation with respect to \(x\): \[ \frac{d}{dx}(4x^3) + \frac{d}{dx}(x^2y) - \frac{d}{dx}(2y^3) = 0. \] This gives: \[ 12x^2 + (2xy + x^2\frac{dy}{dx}) - 6y^2\frac{dy}{dx} = 0. \] Rearranging, we find: \[ x^2\frac{dy}{dx} - 6y^2\frac{dy}{dx} = -12x^2 - 2xy. \] Factoring out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(x^2 - 6y^2) = -12x^2 - 2xy. \] Finally, solving for \(\frac{dy}{dx}\), we arrive at: \[ \frac{dy}{dx} = \frac{-12x^2 - 2xy}{x^2 - 6y^2}. \] Now for \(F_x\): To find \(F_x\), take the partial derivative of \(F(x, y)\): \[ F_x = \frac{\partial}{\partial x}(4x^3 + x^2y - 2y^3) = 12x^2 + 2xy. \]

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If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.

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If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.
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