check whether f(x,y)=root(mod (xy+2y-3x-6)) is differentiable at (x,y)=(2,-3)

To determine whether the function \( f(x, y) = \sqrt{|xy + 2y - 3x - 6|} \) is differentiable at the point \( (2, -3) \), we need to analyze the function's behavior around that point. Differentiability requires that the function is continuous and that its partial derivatives exist and are continuous in a neighborhood around the point of interest.

Step 1: Evaluate the Function at the Point

First, let's substitute \( x = 2 \) and \( y = -3 \) into the function:

\[ f(2, -3) = \sqrt{|(2)(-3) + 2(-3) - 3(2) - 6|} \] \[ = \sqrt{| -6 - 6 - 6 - 6 |} \] \[ = \sqrt{|-24|} = \sqrt{24} = 2\sqrt{6}. \]

Step 2: Check for Continuity

Next, we need to check if the function is continuous at \( (2, -3) \). For continuity, we need to ensure that the limit of \( f(x, y) \) as \( (x, y) \) approaches \( (2, -3) \) equals \( f(2, -3) \).

We can analyze the expression inside the absolute value:

\[ xy + 2y - 3x - 6 = (2)(-3) + 2(-3) - 3(2) - 6 = -24. \]

Since the expression evaluates to a negative number, the absolute value will affect the function. However, as we approach \( (2, -3) \), the expression \( xy + 2y - 3x - 6 \) will remain negative in a neighborhood around this point, ensuring that the absolute value does not introduce any discontinuities.

Step 3: Calculate Partial Derivatives

Now, we need to find the partial derivatives of \( f \) with respect to \( x \) and \( y \). The partial derivatives will help us understand the behavior of the function around the point.

Partial Derivative with Respect to x

Using the chain rule, we have:

\[ f_x(x, y) = \frac{1}{2\sqrt{|xy + 2y - 3x - 6|}} \cdot \frac{\partial}{\partial x} |xy + 2y - 3x - 6|. \]

Calculating the derivative inside the absolute value:

\[ \frac{\partial}{\partial x}(xy + 2y - 3x - 6) = y - 3. \]

At the point \( (2, -3) \), this becomes:

\[ f_x(2, -3) = \frac{1}{2\sqrt{24}} \cdot (-3 - 3) = \frac{-6}{2\sqrt{24}} = \frac{-3}{\sqrt{24}}. \]

Partial Derivative with Respect to y

Similarly, for the partial derivative with respect to \( y \):

\[ f_y(x, y) = \frac{1}{2\sqrt{|xy + 2y - 3x - 6|}} \cdot \frac{\partial}{\partial y} |xy + 2y - 3x - 6|. \]

Calculating the derivative inside the absolute value:

\[ \frac{\partial}{\partial y}(xy + 2y - 3x - 6) = x + 2. \]

At the point \( (2, -3) \), this becomes:

\[ f_y(2, -3) = \frac{1}{2\sqrt{24}} \cdot (2 + 2) = \frac{4}{2\sqrt{24}} = \frac{2}{\sqrt{24}}. \]

Step 4: Check Continuity of Partial Derivatives

Both partial derivatives exist at the point \( (2, -3) \) and are continuous in a neighborhood around that point since they are derived from polynomial functions, which are continuous everywhere.

Final Thoughts

Since \( f(x, y) \) is continuous at \( (2, -3) \) and both partial derivatives exist and are continuous, we can conclude that the function \( f(x, y) \) is differentiable at the point \( (2, -3) \).