Problem 27 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$f(x, y)=x^{2} y^{3}$$

Step-by-Step Solution

Answer

Question: Find the four second partial derivatives of the function \(f(x, y) = x^{2} y^{3}\). Answer: The four second partial derivatives of the function are \(f_{xx} = 2y^3\), \(f_{xy} = 6xy^2\), \(f_{yx} = 6xy^2\), and \(f_{yy} = 6x^2 y\).

Step 1: Find the first partial derivatives

To find the first partial derivatives, differentiate the function with respect to x and y. Let's denote them as \(f_x\) and \(f_y\) respectively. $$f_x = \frac{\partial f(x,y)}{\partial x} = 2x y^3$$ $$f_y = \frac{\partial f(x,y)}{\partial y} = 3x^2 y^2$$

Step 2: Find the second partial derivatives

Now we'll differentiate the first partial derivatives we found in Step 1 with respect to both x and y for each of them. We will calculate \(f_{xx}\), \(f_{xy}\), \(f_{yx}\), \(f_{yy}\), which are the four second partial derivatives we need. $$f_{xx} = \frac{\partial^2 f(x,y)}{\partial x^2} = \frac{\partial f_x}{\partial x} = 2y^3$$ $$f_{xy} = \frac{\partial^2 f(x,y)}{\partial x \partial y} = \frac{\partial f_x}{\partial y} = 6xy^2$$ $$f_{yx} = \frac{\partial^2 f(x,y)}{\partial y \partial x} = \frac{\partial f_y}{\partial x} = 6xy^2$$ $$f_{yy} = \frac{\partial^2 f(x,y)}{\partial y^2} = \frac{\partial f_y}{\partial y} = 6x^2 y$$ So, the four second partial derivatives of the function \(f(x, y) = x^{2} y^{3}\) are: $$f_{xx} = 2y^3,\ f_{xy} = 6xy^2,\ f_{yx} = 6xy^2,\ f_{yy} = 6x^2 y$$