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ind the following derivatives. \( z_{s} \) and \( z_{\text {}} \), where \( z=e^{3 x+5 y}, x=8 s t \), and \( y=6 s+5 t \) \( \frac{\partial z}{\partial x}=3 e^{3 x+5 y} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial x}{\partial s}=8 t \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial z}{\partial y}=5 e^{3 x+5 y} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial y}{\partial s}=\square \) (Type an expression using \( s \) and \( t \) as the variables.)

Ask by Moran Weaver. in the United States
Mar 17,2025

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\( \frac{\partial z}{\partial x} = 3e^{3x+5y} \), \( \frac{\partial x}{\partial s} = 8t \), \( \frac{\partial z}{\partial y} = 5e^{3x+5y} \), \( \frac{\partial y}{\partial s} = 6 \)

Solución

To find the derivatives \( \frac{\partial z}{\partial x} \), \( \frac{\partial x}{\partial s} \), \( \frac{\partial z}{\partial y} \), and \( \frac{\partial y}{\partial s} \), we will use the chain rule and the given expressions for \( z \), \( x \), and \( y \). Given: \( z = e^{3x+5y} \) \( x = 8st \) \( y = 6s + 5t \) 1. Find \( \frac{\partial z}{\partial x} \): Using the chain rule, we have: \( \frac{\partial z}{\partial x} = 3e^{3x+5y} \) 2. Find \( \frac{\partial x}{\partial s} \): Differentiating \( x = 8st \) with respect to \( s \), we get: \( \frac{\partial x}{\partial s} = 8t \) 3. Find \( \frac{\partial z}{\partial y} \): Using the chain rule, we have: \( \frac{\partial z}{\partial y} = 5e^{3x+5y} \) 4. Find \( \frac{\partial y}{\partial s} \): Differentiating \( y = 6s + 5t \) with respect to \( s \), we get: \( \frac{\partial y}{\partial s} = 6 \) Therefore, the derivatives are: \( \frac{\partial z}{\partial x} = 3e^{3x+5y} \) \( \frac{\partial x}{\partial s} = 8t \) \( \frac{\partial z}{\partial y} = 5e^{3x+5y} \) \( \frac{\partial y}{\partial s} = 6 \)

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The derivative \( \frac{\partial y}{\partial s} \) can be found from the expression for \( y \): Given that \( y = 6s + 5t \), the derivative with respect to \( s \) is: \[ \frac{\partial y}{\partial s} = 6 \] Now let's summarize the derivatives you've found: 1. \( \frac{\partial z}{\partial x} = 3 e^{3x + 5y} \) 2. \( \frac{\partial x}{\partial s} = 8t \) 3. \( \frac{\partial z}{\partial y} = 5 e^{3x + 5y} \) 4. \( \frac{\partial y}{\partial s} = 6 \)

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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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