The problem asks us to verify that the mixed partial derivatives are equal, i.e., $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$ for the given functions. We will solve problem 17. $f(x, y) = 2x^2y^3 - x^3y^5$.

1. Problem Description

The problem asks us to verify that the mixed partial derivatives are equal, i.e., $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$ for the given functions. We will solve problem

1

7. $f(x, y) = 2x^2y^3 - x^3y^5$.

2. Solution Steps

First, we find the partial derivative of $f$ with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (2x^2y^3 - x^3y^5) = 4xy^3 - 3x^2y^5$

Next, we find the partial derivative of $\frac{\partial f}{\partial x}$ with respect to $y$:

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (4xy^3 - 3x^2y^5) = 12xy^2 - 15x^2y^4$

Now, we find the partial derivative of $f$ with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (2x^2y^3 - x^3y^5) = 6x^2y^2 - 5x^3y^4$

Finally, we find the partial derivative of $\frac{\partial f}{\partial y}$ with respect to $x$:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (6x^2y^2 - 5x^3y^4) = 12xy^2 - 15x^2y^4$

Since $\frac{\partial^2 f}{\partial y \partial x} = 12xy^2 - 15x^2y^4$ and $\frac{\partial^2 f}{\partial x \partial y} = 12xy^2 - 15x^2y^4$, we have verified that $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$.

3. Final Answer

We have verified that $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$ for $f(x, y) = 2x^2y^3 - x^3y^5$.

Both mixed partial derivatives are $12xy^2 - 15x^2y^4$.