Problem 28 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The second partial derivatives of the function $$f(x, y) = (x + 3y)^2$$ are: 1. $$\frac{\partial^2f}{\partial x^2} = 2$$ 2. $$\frac{\partial^2f}{\partial y^2} = 18$$ 3. $$\frac{\partial^2f}{\partial x \partial y} = 6$$ 4. $$\frac{\partial^2f}{\partial y \partial x} = 6$$

Step 1: Find the first partial derivatives

To find the first partial derivatives, we will differentiate the function with respect to x, and then with respect to y. $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x + 3y)^2 = 2(x + 3y)\frac{\partial}{\partial x}(x + 3y) = 2(x + 3y)$$ $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x + 3y)^2 = 2(x + 3y)\frac{\partial}{\partial y}(x + 3y) = 6(x + 3y)$$

Step 2: Find the second partial derivatives

Now, we will differentiate these first partial derivatives again with respect to x and y to find the second partial derivatives. 1. $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(2(x + 3y)\right) = 2$$ 2. $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(6(x + 3y)\right) = 18$$ 3. $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}\left(6(x + 3y)\right) = 6$$ 4. $$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}\left(2(x + 3y)\right) = 6$$ So, the four second partial derivatives are: 1. $$\frac{\partial^2f}{\partial x^2} = 2$$ 2. $$\frac{\partial^2f}{\partial y^2} = 18$$ 3. $$\frac{\partial^2f}{\partial x \partial y} = 6$$ 4. $$\frac{\partial^2f}{\partial y \partial x} = 6$$