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Find the derivative \( \frac{d w}{d t} \), where \( w=6 x y z, x=4 t^{4}, y=t^{-1} \), and \( z=2 t^{-3} \) \( \frac{\partial w}{\partial x}=6 y z \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{\partial w}{\partial y}=6 x z \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{\partial w}{\partial z}=6 x y \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{d x}{d t}=\square \) (Type an expression using \( t \) as the variable.)

Ask by Gonzalez Stuart. in the United States
Mar 17,2025

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\[ \frac{dx}{dt} = 16t^3 \]

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To begin, let's express \( w \) in a simpler form: Given \( w = 6xyz \), we know that \( x = 4t^4 \), \( y = t^{-1} \), and \( z = 2t^{-3} \). Now, let's calculate \( \frac{d w}{d t} \) using the chain rule: \[ \frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \] With the given partial derivatives: 1. \( \frac{\partial w}{\partial x} = 6yz \) 2. \( \frac{\partial w}{\partial y} = 6xz \) 3. \( \frac{\partial w}{\partial z} = 6xy \) Next, we need the derivatives of \( x, y, z \) with respect to \( t \): - Since \( x = 4t^{4} \), we have \( \frac{dx}{dt} = 16t^{3} \). - For \( y = t^{-1} \), \( \frac{dy}{dt} = -t^{-2} \). - For \( z = 2t^{-3} \), \( \frac{dz}{dt} = -6t^{-4} \). Now substituting everything back into the expression for \( \frac{d w}{dt}\): \[ \frac{d w}{d t} = 6yz \cdot 16t^{3} + 6xz \cdot (-t^{-2}) + 6xy \cdot (-6t^{-4}) \] Inserting the expressions for \( x, y, z \): 1. \( \frac{dx}{dt} = 16t^{3} \) 2. \( \frac{dy}{dt} = -t^{-2} \) 3. \( \frac{dz}{dt} = -6t^{-4} \) This will give you the overall \( \frac{d w}{d t} \) depending on \( t \) itself. Thus we conclude: - \( \frac{dx}{dt} = 16t^{3} \) - \( \frac{dy}{dt} = -t^{-2} \) - \( \frac{dz}{dt} = -6t^{-4} \) Use these derivatives accordingly to find the final expression for \( \frac{d w}{d t} \).

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