To solve this problem, we need to compute the first-order partial derivatives of the given function $f(x, y) = (x^2 + 1) \ln(xy^2)$.

Step 1: Expand the function $f(x, y)$

The function is given as:

$f(x, y) = (x^2 + 1) \ln(xy^2)$

We can use the logarithmic property $\ln(ab) = \ln(a) + \ln(b)$ to expand:

$f(x, y) = (x^2 + 1) \ln(x) + (x^2 + 1) \ln(y^2)$

Since $\ln(y^2) = 2 \ln(y)$, the expression becomes:

$f(x, y) = (x^2 + 1) \ln(x) + 2(x^2 + 1) \ln(y)$

Step 2: Compute the first-order partial derivatives

Partial derivative with respect to $x$

We need to differentiate $f(x, y)$ with respect to $x$. Using the product rule:

$\frac{\partial}{\partial x} \left[ (x^2 + 1) \ln(x) \right] = 2x \ln(x) + \frac{x^2 + 1}{x}$

Now for the second term:

$\frac{\partial}{\partial x} \left[ 2(x^2 + 1) \ln(y) \right] = 4x \ln(y)$

Thus, the total partial derivative with respect to $x$ is:

$f_x = 2x \ln(x) + \frac{x^2 + 1}{x} + 4x \ln(y)$

Partial derivative with respect to $y$

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

$\frac{\partial}{\partial y} \left[ (x^2 + 1) \ln(x) \right] = 0$

$\frac{\partial}{\partial y} \left[ 2(x^2 + 1) \ln(y) \right] = \frac{2(x^2 + 1)}{y}$

So the total partial derivative with respect to $y$ is:

$f_y = \frac{2(x^2 + 1)}{y}$

Step 3: Verify with the choices

Now, let's compare the partial derivatives we found with the options provided:

Option (a):

$[2x \ln(xy^2) + (x^2 + 1)y^2 \ln(xy^2)] dx + \frac{2(x^2 + 1)}{y} dy$

This looks different from our result, as the terms involving $y^2$ and $\ln(xy^2)$ are not present in our derivative.

Option (b):

$[2x \ln(xy^2) + (x^2 + 1)y^2 \ln(xy^2)] + \frac{2(x^2 + 1)}{y}$

Also doesn't match.

Option (c):

$[2x \ln(xy^2) + \frac{x^2 + 1}{x}] dx + \frac{2(x^2 + 1)}{y} dy$

This matches with our derived partial derivatives:

• $2x \ln(xy^2) + \frac{x^2 + 1}{x}$ corresponds to $f_x$
• $\frac{2(x^2 + 1)}{y}$ corresponds to $f_y$

So, option (c) is the correct answer.

Final Answer:

Option (c) is the correct choice.

Let me know if you need further details or have any questions!

5 Related Questions:

1. How would the partial derivatives change if the function involved a cubic term instead of $x^2 + 1$?
2. How do logarithmic differentiation rules apply to functions with multiple variables?
3. How would we compute the second-order partial derivatives of $f(x, y)$?
4. What is the geometric interpretation of the first-order partial derivatives in this context?
5. How can we use the chain rule in multi-variable functions like $f(x, y)$?

Tip:

When differentiating functions with logarithmic terms, remember to apply the chain rule carefully, especially when multiple variables are involved.