We are tasked with calculating the second mixed partial derivative $f_{xy}(x, y)$ of the function:

$f(x, y) = e^{bx + ay}$
where $b$ and $a$ are constants. Let's compute it step by step.

Step 1: First partial derivative with respect to $x$:

To differentiate $f(x, y) = e^{bx + ay}$ with respect to $x$, treat $y$ as a constant. The derivative of the exponential function is:

$\frac{\partial}{\partial x} e^{bx+ay} = e^{bx+ay} \cdot \frac{\partial}{\partial x}(bx+ay).$

Since $\frac{\partial}{\partial x}(bx + ay) = b$, we get:

$f_x(x, y) = b e^{bx + ay}.$

Step 2: Mixed partial derivative with respect to $y$:

Now differentiate $f_x(x, y)$ with respect to $y$:

$f_x(x, y) = b e^{bx + ay}.$

Taking the partial derivative with respect to $y$:

$\frac{\partial}{\partial y} f_x(x, y) = b \cdot \frac{\partial}{\partial y} e^{bx+ay}.$

The derivative of $e^{bx+ay}$ with respect to $y$ is:

$\frac{\partial}{\partial y} e^{bx+ay} = e^{bx+ay} \cdot \frac{\partial}{\partial y}(bx + ay).$

Since $\frac{\partial}{\partial y}(bx + ay) = a$, we get:

$\frac{\partial}{\partial y} e^{bx+ay} = a e^{bx+ay}.$

Thus:

$f_{xy}(x, y) = b \cdot a e^{bx + ay}.$

Final Answer:

The second mixed partial derivative $f_{xy}(x, y)$ is:

$f_{xy}(x, y) = ab e^{bx + ay}.$

Do you want further clarification or steps for verification? Let me know!

Here are 5 related questions for practice:

1. Compute $f_{xx}(x, y)$ for the same function $f(x, y) = e^{bx+ay}$.
2. Compute $f_{yy}(x, y)$ for the same function.
3. For $f(x, y) = e^{x^2 + y^2}$, find $f_{xy}$.
4. Prove that the mixed partial derivatives $f_{xy}$ and $f_{yx}$ are equal for this function.
5. Differentiate $f(x, y) = e^{bxy}$ and find $f_{xy}(x, y)$.

Tip:

When working with exponential functions, remember to use the chain rule carefully, as constants and exponents play a significant role in differentiation!