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Let's break down the given problem step by step and solve the requested parts.

Given that \( A = 5 \) and \( B = 6 \), the function \( f : \mathbb{R}^2 \to \mathbb{R} \) is defined as:

\[
f(x, y) = 
\begin{cases} 
\frac{(A+1) x^3 y - (B+1) x y^3}{x^2 + y^2}, & \text{if } (x, y) \neq (0, 0), \\
0, & \text{if } (x, y) = (0, 0).
\end{cases}
\]

Substituting \( A = 5 \) and \( B = 6 \), the function becomes:

\[
f(x, y) = 
\begin{cases} 
\frac{6 x^3 y - 7 x y^3}{x^2 + y^2}, & \text{if } (x, y) \neq (0, 0), \\
0, & \text{if } (x, y) = (0, 0).
\end{cases}
\]

Now, let’s address each part of the problem:

### (a) [20 points] Find the formula for \( f_1(x, y) \) and \( f_2(x, y) \) for \( (x, y) \neq (0, 0) \).

The partial derivatives \( f_1(x, y) \) and \( f_2(x, y) \) represent the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), respectively.

#### Step 1: Compute \( f_1(x, y) \), the partial derivative with respect to \( x \).
We'll differentiate the function:

\[
f(x, y) = \frac{6 x^3 y - 7 x y^3}{x^2 + y^2}
\]

Using the quotient rule:

\[
f_1(x, y) = \frac{(x^2 + y^2)\frac{\partial}{\partial x} (6x^3 y - 7 x y^3) - (6 x^3 y - 7 x y^3) \frac{\partial}{\partial x} (x^2 + y^2)}{(x^2 + y^2)^2}
\]

Now, we compute the necessary derivatives:
\[
\frac{\partial}{\partial x} (6x^3 y - 7xy^3) = 18x^2 y - 7 y^3
\]
and
\[
\frac{\partial}{\partial x} (x^2 + y^2) = 2x.
\]

Thus,
\[
f_1(x, y) = \frac{(x^2 + y^2)(18x^2 y - 7 y^3) - (6x^3 y - 7 x y^3)(2x)}{(x^2 + y^2)^2}.
\]

#### Step 2: Compute \( f_2(x, y) \), the partial derivative with respect to \( y \).

Similarly, we compute the partial derivative of \( f(x, y) \) with respect to \( y \):

\[
f_2(x, y) = \frac{(x^2 + y^2)\frac{\partial}{\partial y} (6x^3 y - 7 x y^3) - (6 x^3 y - 7 x y^3) \frac{\partial}{\partial y} (x^2 + y^2)}{(x^2 + y^2)^2}
\]

The derivatives are:
\[
\frac{\partial}{\partial y} (6x^3 y - 7xy^3) = 6x^3 - 21xy^2
\]
and
\[
\frac{\partial}{\partial y} (x^2 + y^2) = 2y.
\]

Thus,
\[
f_2(x, y) = \frac{(x^2 + y^2)(6x^3 - 21xy^2) - (6x^3 y - 7xy^3)(2y)}{(x^2 + y^2)^2}.
\]

### (b) [15 points] Show that \( f_1(0, 0) = f_2(0, 0) = 0 \).

We need to check the behavior of the partial derivatives at \( (x, y) = (0, 0) \). Since the function \( f(x, y) \) is defined to be \( 0 \) at \( (0, 0) \), we expect that both partial derivatives will also be \( 0 \) at this point. Let's verify this by taking the limits.

\[
f_1(0, 0) = \lim_{(x, y) \to (0, 0)} \frac{(x^2 + y^2)(18x^2 y - 7 y^3) - (6x^3 y - 7xy^3)(2x)}{(x^2 + y^2)^2}
\]

We observe that both the numerator and denominator go to zero as \( (x, y) \to (0, 0) \). Using polar coordinates or L'Hôpital's rule, we can show that this limit is \( 0 \). Similarly, for \( f_2(

Misalkan AB menyatakan dua digit terakhir NIU Saudara. Sebagai contoh, jika NIU Saudara adalah 123456, maka A = 5 dan B = 6. Diberikan fungsi f: ℝ^2 → ℝ dengan f(x, y) = {( (A+1)x^3y - (B+1)xy^3 )/(x^2 + y^2) jika (x, y) ≠ (0,0), 0 jika (x, y) = (0,0). Tentukan f1(x,y) dan f2(x,y) serta tunjukkan bahwa f1(0,0) = f2(0,0). Selidiki juga apakah f12(0,0) dan f21(0,0) ada, dan apakah f diferensiabel di (0,0).

Solve a multivariable calculus problem involving partial derivatives and differentiability. The problem examines a function f(x, y) defined using constants A and B. Learn how to find f1(x, y), f2(x, y), and analyze the function's behavior at (0,0). Suitable for undergraduate students.

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