Problem 25 - Calculus Early Transcendentals

Question

Find the four second partial derivatives of the following functions. $$h(x, y)=x^{3}+x y^{2}+1$$

Step-by-Step Solution

Answer

Answer: The four second-order partial derivatives are: $$ \frac{\partial^2 h}{\partial x^2} = 6x, \frac{\partial^2h}{\partial x \partial y} = 2y, \frac{\partial^2h}{\partial y \partial x} = 2y, \text{ and } \frac{\partial^2h}{\partial y^2} = 0 $$

Step 1: Find the first-order partial derivatives of h(x, y)

To find the first-order partial derivatives, we'll differentiate the function h(x, y) with respect to x and y: $$ \frac{\partial h}{\partial x} = \frac{\partial (x^3 + xy^2 + 1)}{\partial x} $$ and $$ \frac{\partial h}{\partial y} = \frac{\partial (x^3 + xy^2 + 1)}{\partial y} $$

Step 2: Calculate the first-order partial derivatives

Differentiate h(x, y) with respect to x and y: $$ \frac{\partial h}{\partial x} = 3x^2 + y^2 $$ and $$ \frac{\partial h}{\partial y} = 2xy $$

Step 3: Find the second-order partial derivatives

Now, we'll differentiate the first-order partial derivatives obtained in step 2 with respect to x and y. This will give us the four second-order partial derivatives: $$ \frac{\partial^2 h}{\partial x^2} = \frac{\partial (\frac{\partial h}{\partial x})}{\partial x} $$ $$ \frac{\partial^2 h}{\partial x \partial y} = \frac{\partial (\frac{\partial h}{\partial x})}{\partial y} $$ $$ \frac{\partial^2 h}{\partial y \partial x} = \frac{\partial (\frac{\partial h}{\partial y})}{\partial x} $$ $$ \frac{\partial^2 h}{\partial y^2} = \frac{\partial (\frac{\partial h}{\partial y})}{\partial y} $$

Step 4: Calculate the second-order partial derivatives

Differentiate the first-order partial derivatives with respect to x and y to obtain the second-order partial derivatives: $$ \frac{\partial^2 h}{\partial x^2} = 6x $$ $$ \frac{\partial^2h}{\partial x \partial y} = 2y $$ $$ \frac{\partial^2h}{\partial y \partial x} = 2y $$ $$ \frac{\partial^2h}{\partial y^2}=0 $$ The four second-order partial derivatives are: $$ \frac{\partial^2 h}{\partial x^2} = 6x, \frac{\partial^2h}{\partial x \partial y} = 2y, \frac{\partial^2h}{\partial y \partial x} = 2y, \text{ and } \frac{\partial^2h}{\partial y^2} = 0 $$