Find the first partial derivatives of the function: $u = t e^{\frac{2w}{t}}.$

Neetesh Kumar | December 3, 2024

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This is the solution to Math 1D
Assignment: 14.3 Question Number 15
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Step-by-step solution:

Step 1: Partial derivative of $u$ with respect to $t$

To find $\frac{\partial u}{\partial t}$, treat $w$ as constant and differentiate $u$ with respect to $t$: $u = t e^{\frac{2w}{t}}.$

Using the product rule: $\frac{\partial u}{\partial t} = \frac{\partial}{\partial t}\left(t\right)e^{\frac{2w}{t}} + t \frac{\partial}{\partial t}\left(e^{\frac{2w}{t}}\right).$

1. For the first term: $\frac{\partial}{\partial t}\left(t\right)e^{\frac{2w}{t}} = e^{\frac{2w}{t}}.$

2. For the second term, differentiate $e^{\frac{2w}{t}}$ using the chain rule: $\frac{\partial}{\partial t}\left(e^{\frac{2w}{t}}\right) = e^{\frac{2w}{t}} \cdot \frac{\partial}{\partial t}\left(\frac{2w}{t}\right).$

The derivative of $\frac{2w}{t}$ with respect to $t$ is: $\frac{\partial}{\partial t}\left(\frac{2w}{t}\right) = -\frac{2w}{t^2}.$

Thus: $\frac{\partial}{\partial t}\left(e^{\frac{2w}{t}}\right) = e^{\frac{2w}{t}} \cdot \left(-\frac{2w}{t^2}\right).$

Combine both terms: $\frac{\partial u}{\partial t} = e^{\frac{2w}{t}} - \frac{2w}{t}e^{\frac{2w}{t}}.$

Factorize: $\frac{\partial u}{\partial t} = e^{\frac{2w}{t}}\left(1 - \frac{2w}{t}\right).$

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

To find $\frac{\partial u}{\partial w}$, treat $t$ as constant and differentiate $u$ with respect to $w$: $u = t e^{\frac{2w}{t}}.$

Differentiate with respect to $w$: $\frac{\partial u}{\partial w} = t \cdot \frac{\partial}{\partial w}\left(e^{\frac{2w}{t}}\right).$

Using the chain rule: $\frac{\partial}{\partial w}\left(e^{\frac{2w}{t}}\right) = e^{\frac{2w}{t}} \cdot \frac{\partial}{\partial w}\left(\frac{2w}{t}\right).$

The derivative of $\frac{2w}{t}$ with respect to $w$ is: $\frac{\partial}{\partial w}\left(\frac{2w}{t}\right) = \frac{2}{t}.$

Thus: $\frac{\partial u}{\partial w} = t \cdot e^{\frac{2w}{t}} \cdot \frac{2}{t}.$

Simplify: $\frac{\partial u}{\partial w} = 2 e^{\frac{2w}{t}}.$

Final Answers:

$\frac{\partial u}{\partial t} = \boxed{e^{\frac{2w}{t}}\left(1 - \frac{2w}{t}\right)}$

$\frac{\partial u}{\partial w} = \boxed{2 e^{\frac{2w}{t}}}$

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