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For the function f(x,y)= x + 4y 3 4 x +y find f, fy, f(5,-5), and f,(1,3).

For the function f(x,y)= x + 4y 3 4 x +y find f, fy, f(5,-5), and f,(1,3).

Calculus: Early Transcendentals
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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### Calculating Partial Derivatives for a Given Function For the function \( f(x, y) = \frac{x^2 + 4y^3}{\frac{3}{4}x + y} \), follow these steps to find the partial derivatives \( f_x \), \( f_y \), as well as evaluate \( f_x(5, -5) \) and \( f_y(1, 3) \): 1. **Find the Partial Derivative with Respect to \( x \), \( f_x \):** - Begin by applying the quotient rule for differentiation, where \( u(x, y) = x^2 + 4y^3 \) and \( v(x, y) = \frac{3}{4}x + y \). - The quotient rule states that \( \left( \frac{u}{v} \right)_x = \frac{u_x v - u v_x}{v^2} \). Calculate \( u_x \): \( u_x = \frac{\partial}{\partial x} (x^2 + 4y^3) = 2x \) Calculate \( v_x \): \( v_x = \frac{\partial}{\partial x} \left(\frac{3}{4}x + y\right) = \frac{3}{4} \) Therefore, \( f_x = \frac{(2x)(\frac{3}{4}x + y) - (x^2 + 4y^3)(\frac{3}{4})}{\left(\frac{3}{4}x + y\right)^2} \) 2. **Find the Partial Derivative with Respect to \( y \), \( f_y \):** - Similarly, use the quotient rule where \( u(x, y) = x^2 + 4y^3 \) and \( v(x, y) = \frac{3}{4}x + y \). - The quotient rule for \( y \) states that \( \left( \frac{u}{v} \right)_y = \frac{u_y v - u v_y}{v^2} \). Calculate \( u_y \): \( u_y = \frac{\partial}{\partial y} (x^2 + 4y^3) =
Transcribed Image Text:### Calculating Partial Derivatives for a Given Function For the function \( f(x, y) = \frac{x^2 + 4y^3}{\frac{3}{4}x + y} \), follow these steps to find the partial derivatives \( f_x \), \( f_y \), as well as evaluate \( f_x(5, -5) \) and \( f_y(1, 3) \): 1. **Find the Partial Derivative with Respect to \( x \), \( f_x \):** - Begin by applying the quotient rule for differentiation, where \( u(x, y) = x^2 + 4y^3 \) and \( v(x, y) = \frac{3}{4}x + y \). - The quotient rule states that \( \left( \frac{u}{v} \right)_x = \frac{u_x v - u v_x}{v^2} \). Calculate \( u_x \): \( u_x = \frac{\partial}{\partial x} (x^2 + 4y^3) = 2x \) Calculate \( v_x \): \( v_x = \frac{\partial}{\partial x} \left(\frac{3}{4}x + y\right) = \frac{3}{4} \) Therefore, \( f_x = \frac{(2x)(\frac{3}{4}x + y) - (x^2 + 4y^3)(\frac{3}{4})}{\left(\frac{3}{4}x + y\right)^2} \) 2. **Find the Partial Derivative with Respect to \( y \), \( f_y \):** - Similarly, use the quotient rule where \( u(x, y) = x^2 + 4y^3 \) and \( v(x, y) = \frac{3}{4}x + y \). - The quotient rule for \( y \) states that \( \left( \frac{u}{v} \right)_y = \frac{u_y v - u v_y}{v^2} \). Calculate \( u_y \): \( u_y = \frac{\partial}{\partial y} (x^2 + 4y^3) =
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