Find the first partial derivatives of the function: $w = 6ze^{xyz}$

Neetesh Kumar | December 3, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.3 Question Number 17
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Step-by-step solution:

Step 1: Partial derivative of $w$ with respect to $x$

To compute $\frac{\partial w}{\partial x}$, treat $y$ and $z$ as constants. Differentiate $w$ with respect to $x$:

Using the chain rule: $\frac{\partial w}{\partial x} = 6z \cdot \frac{\partial}{\partial x}\left(e^{xyz}\right).$

For $e^{xyz}$: $\frac{\partial}{\partial x}\left(e^{xyz}\right) = e^{xyz} \cdot \frac{\partial}{\partial x}\left(xyz\right) = e^{xyz} \cdot yz.$

Thus: $\frac{\partial w}{\partial x} = 6z \cdot e^{xyz} \cdot yz.$

Simplify: $\frac{\partial w}{\partial x} = 6yz^2e^{xyz}.$

Step 2: Partial derivative of $w$ with respect to $y$

To compute $\frac{\partial w}{\partial y}$, treat $x$ and $z$ as constants. Differentiate $w$ with respect to $y$:

Using the chain rule: $\frac{\partial w}{\partial y} = 6z \cdot \frac{\partial}{\partial y}\left(e^{xyz}\right).$

For $e^{xyz}$: $\frac{\partial}{\partial y}\left(e^{xyz}\right) = e^{xyz} \cdot \frac{\partial}{\partial y}\left(xyz\right) = e^{xyz} \cdot xz.$

Thus: $\frac{\partial w}{\partial y} = 6z \cdot e^{xyz} \cdot xz.$

Simplify: $\frac{\partial w}{\partial y} = 6xz^2e^{xyz}.$

Step 3: Partial derivative of $w$ with respect to $z$

To compute $\frac{\partial w}{\partial z}$, treat $x$ and $y$ as constants. Differentiate $w$ with respect to $z$:

Using the product rule: $\frac{\partial w}{\partial z} = \frac{\partial}{\partial z}\left(6z\right)e^{xyz} + 6z \cdot \frac{\partial}{\partial z}\left(e^{xyz}\right).$

1. Differentiate the first term: $\frac{\partial}{\partial z}\left(6z\right)e^{xyz} = 6e^{xyz}.$

2. For the second term, differentiate $e^{xyz}$: $\frac{\partial}{\partial z}\left(e^{xyz}\right) = e^{xyz} \cdot \frac{\partial}{\partial z}\left(xyz\right) = e^{xyz} \cdot xy.$

Thus: $\frac{\partial w}{\partial z} = 6e^{xyz} + 6z \cdot e^{xyz} \cdot xy.$

Simplify: $\frac{\partial w}{\partial z} = 6e^{xyz} \left(1 + xyz\right).$

Final Answers:

$\frac{\partial w}{\partial x} = \boxed{6yz^2e^{xyz}}$

$\frac{\partial w}{\partial y} = \boxed{6xz^2e^{xyz}}$

$\frac{\partial w}{\partial z} = \boxed{6e^{xyz} \left(1 + xyz\right)}$

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