Use the Chain Rule to evaluate the partial derivative at the point (r, 0) = (2V2, ), where g(x, y) : -, x = 8r cos(0), x+y y = 9r sin(0). (Use symbolic notation and fractions where needed.) dg (r.0)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Title: Evaluating Partial Derivatives Using the Chain Rule**

**Objective:**
Learn how to evaluate the partial derivative of a function using the chain rule in the context of polar coordinates.

**Problem Statement:**
Use the Chain Rule to evaluate the partial derivative \(\frac{\partial g}{\partial \theta}\) at the point \((r, \theta) = \left(2\sqrt{2}, \frac{\pi}{4}\right)\), where 

- \(g(x, y) = \frac{1}{x+y^2}\),
- \(x = 8r \cos(\theta)\),
- \(y = 9r \sin(\theta)\).

**Instructions:**
Use symbolic notation and fractions where needed to compute the solution.

**Solution:**
Find \(\frac{\partial g}{\partial \theta}\) using the chain rule and express your answer in the box provided:

\[
\frac{\partial g}{\partial \theta} \bigg|_{(r, \theta)} = \boxed{}
\]
Transcribed Image Text:**Title: Evaluating Partial Derivatives Using the Chain Rule** **Objective:** Learn how to evaluate the partial derivative of a function using the chain rule in the context of polar coordinates. **Problem Statement:** Use the Chain Rule to evaluate the partial derivative \(\frac{\partial g}{\partial \theta}\) at the point \((r, \theta) = \left(2\sqrt{2}, \frac{\pi}{4}\right)\), where - \(g(x, y) = \frac{1}{x+y^2}\), - \(x = 8r \cos(\theta)\), - \(y = 9r \sin(\theta)\). **Instructions:** Use symbolic notation and fractions where needed to compute the solution. **Solution:** Find \(\frac{\partial g}{\partial \theta}\) using the chain rule and express your answer in the box provided: \[ \frac{\partial g}{\partial \theta} \bigg|_{(r, \theta)} = \boxed{} \]
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