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Find the total differential. \[ z=e^{-x} \tan (y) \]

Ask by Brooks Wells. in the United States
Mar 16,2025

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The total differential \( dz \) is: \[ dz = -e^{-x}\tan(y)\,dx + e^{-x}\sec^2(y)\,dy \]

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To find the total differential of the function \( z = e^{-x} \tan(y) \), we'll use the formula for the total differential: \[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy \] First, we compute the partial derivatives: 1. **Partial derivative with respect to \( x \)**: \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(e^{-x} \tan(y)) = -e^{-x} \tan(y) \] 2. **Partial derivative with respect to \( y \)**: \[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(e^{-x} \tan(y)) = e^{-x} \sec^2(y) \] Now substituting these into the total differential formula: \[ dz = (-e^{-x} \tan(y)) \, dx + (e^{-x} \sec^2(y)) \, dy \] Combining the terms gives us: \[ dz = -e^{-x} \tan(y) \, dx + e^{-x} \sec^2(y) \, dy \] Therefore, the total differential is: \[ dz = -e^{-x} \tan(y) \, dx + e^{-x} \sec^2(y) \, dy \]

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