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giải bài tập này và chọn đáp đúng theo trong ảnh. Chú ý giải theo chương trình toán 3 bậc đại học

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.

This problem explores the first-order partial derivatives of a multivariable function, f(x, y) = (x^2 + 1) ln(xy^2), including the use of logarithmic properties and the product rule. Step-by-step solution provided. Suitable for university-level calculus students.

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