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Find the following derivatives. \( \mathrm{z}_{\mathrm{s}} \) and \( \mathrm{z}_{\mathrm{L}} \), where \( \mathrm{z}=e^{3 \mathrm{x}+5 \mathrm{y}}, \mathrm{x}=8 \mathrm{st} \), and \( \mathrm{y}=6 \mathrm{~s}+5 \mathrm{t} \) \[ \overline{\partial \mathrm{s}}=8 \mathrm{t} \] (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.) \[ \frac{\partial y}{\partial s}=\overline{6} \] (Type an expression using \( s \) and \( t \) as the variables.) \[ \frac{\partial x}{\partial t}=8 s \] (Type an expression using s and t as the variables.) \[ \frac{\partial y}{\partial t}=5 \] (Type an expression using s and \( t \) as the variables.) \[ z_{s}=\square \] (Type an expression using s and t as the variables.)

Ask by Bryant Christensen. in the United States
Mar 17,2025

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\( z_s = (24t + 30) e^{24st + 30s + 25t} \)

Solución

We begin with \[ z = e^{3x+5y} \] with \[ x = 8st \quad \text{and} \quad y = 6s + 5t. \] Using the chain rule, the derivative of \( z \) with respect to \( s \) is given by \[ z_s = \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}. \] Since \[ \frac{\partial z}{\partial x} = 3e^{3x+5y} \quad \text{and} \quad \frac{\partial z}{\partial y} = 5e^{3x+5y}, \] we now compute the required derivatives of \( x \) and \( y \) with respect to \( s \). Given \[ x = 8st, \] differentiating with respect to \( s \) (treating \( t \) as a constant) gives \[ \frac{\partial x}{\partial s} = 8t. \] Similarly, for \[ y = 6s + 5t, \] the derivative with respect to \( s \) is \[ \frac{\partial y}{\partial s} = 6. \] Plugging these into the chain rule expression, we have \[ z_s = \left(3e^{3x+5y}\right)(8t) + \left(5e^{3x+5y}\right)(6). \] This simplifies to \[ z_s = e^{3x+5y}\left(24t + 30\right). \] Substitute back the expressions for \( x \) and \( y \) in terms of \( s \) and \( t \): \[ 3x + 5y = 3(8st) + 5(6s+5t) = 24st + 30s + 25t. \] Thus, \[ z_s = \left(24t + 30\right) e^{24st + 30s + 25t}. \]

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To find \( z_s \), we can use the chain rule. We have: \[ z = e^{3x + 5y} \] Thus, we apply the chain rule: \[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \] Given that \( \frac{\partial z}{\partial x} = 3e^{3x + 5y} \), \( \frac{\partial z}{\partial y} = 5e^{3x + 5y} \), \( \frac{\partial x}{\partial s} = 0 \) (since x does not depend on s), and \( \frac{\partial y}{\partial s} = 6 \), we substitute these into our expression: \[ z_s = 0 + 5 e^{3x + 5y} \cdot 6 = 30 e^{3x + 5y} \] This leads to the expression: \[ z_s = 30 e^{3(8st) + 5(6s + 5t)} \quad (\text{using } x = 8st \text{ and } y = 6s + 5t) \] Next, for \( z_L \), we will seek to derive it using a similar format. The derivative notation \( z_L \) typically denotes \( \frac{\partial z}{\partial l} \) where \( l \) is an independent variable. However, if we take the total derivatives including the relationships we derived: \[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \] Using \( \frac{\partial x}{\partial t} = 8s \) and \( \frac{\partial y}{\partial t} = 5 \): \[ z_L = 3e^{3x + 5y} \cdot 8s + 5e^{3x + 5y} \cdot 5 = (24s + 25) e^{3x + 5y} \] Thus, we provide the expressions as: \[ z_s = 30 e^{24st + 30s + 25t} \] \[ z_L = (24s + 25) e^{24st + 30s + 25t} \]

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