f(x, y) = √√2x² + 5y² fz(0, 1) =

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Below is the transcription of the given mathematical image, which seems to be about a multivariable function and its partial derivative. Here's how it might appear on an educational website:

---

### Multivariable Function and Partial Derivative

**Function Definition:**

\[ f(x, y) = \sqrt{2x^2 + 5y^2} \]

**Partial Derivative Calculation:**

To find the partial derivative of the function with respect to \( x \) at the point \((0,1)\):

\[ f_x(0, 1) = \_\_\_\_\_\_ \]

**Explanation:**

The given function is a multivariable function defined as:

\[ f(x, y) = \sqrt{2x^2 + 5y^2} \]

To compute the partial derivative of \( f \) with respect to \( x \), represented as \( f_x \), we take the derivative of \( f \) with \( x \) while treating \( y \) as a constant. 

Then, we evaluate this derivative at the point \((0,1)\), which means substituting \( x = 0 \) and \( y = 1 \) into the derivative expression.

---

Note: Students or readers are expected to perform the differentiation and substitute the values to find \( f_x(0, 1) \).
Transcribed Image Text:Below is the transcription of the given mathematical image, which seems to be about a multivariable function and its partial derivative. Here's how it might appear on an educational website: --- ### Multivariable Function and Partial Derivative **Function Definition:** \[ f(x, y) = \sqrt{2x^2 + 5y^2} \] **Partial Derivative Calculation:** To find the partial derivative of the function with respect to \( x \) at the point \((0,1)\): \[ f_x(0, 1) = \_\_\_\_\_\_ \] **Explanation:** The given function is a multivariable function defined as: \[ f(x, y) = \sqrt{2x^2 + 5y^2} \] To compute the partial derivative of \( f \) with respect to \( x \), represented as \( f_x \), we take the derivative of \( f \) with \( x \) while treating \( y \) as a constant. Then, we evaluate this derivative at the point \((0,1)\), which means substituting \( x = 0 \) and \( y = 1 \) into the derivative expression. --- Note: Students or readers are expected to perform the differentiation and substitute the values to find \( f_x(0, 1) \).
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