We are given the function:

$F(x, y) = x^3 + \sqrt{y^2(x - 2)} + |x + 5|$

We need to compute the partial derivative of $F$ with respect to $y$, and analyze where the partial derivative is continuous.

Step 1: Compute the Partial Derivative with Respect to $y$

We differentiate each term of the function with respect to $y$, keeping $x$ constant.

1. The term $x^3$ does not depend on $y$, so its derivative with respect to $y$ is $0$.

2. The second term is $\sqrt{y^2(x - 2)}$. Using the chain rule, we differentiate: $\frac{\partial}{\partial y} \left( \sqrt{y^2(x - 2)} \right)$ This becomes: $\frac{1}{2\sqrt{y^2(x - 2)}} \cdot 2y(x - 2) = \frac{y(x - 2)}{\sqrt{y^2(x - 2)}}$

3. The third term is $|x + 5|$, which does not depend on $y$, so its derivative with respect to $y$ is $0$.

Thus, the partial derivative of $F$ with respect to $y$ is:

$\frac{\partial F}{\partial y} = \frac{y(x - 2)}{\sqrt{y^2(x - 2)}}$

Step 2: Analyze Continuity of the Partial Derivative

To determine where the partial derivative is continuous, we look for values of $x$ and $y$ that might cause discontinuities. Specifically, the expression involves a square root and a fraction, both of which can introduce potential issues.

1. The square root in $\sqrt{y^2(x - 2)}$ is valid only if $y^2(x - 2) \geq 0$. This means:

   • If $x - 2 > 0$, then $y^2 \geq 0$ is always true.
   • If $x - 2 < 0$, the expression is undefined unless $y = 0$.
   • If $x - 2 = 0$, the expression is $\sqrt{0} = 0$, which is well-defined for any $y$.

2. The denominator of the fraction $\sqrt{y^2(x - 2)}$ also introduces a potential issue if it is zero. Specifically, the denominator is zero when $y = 0$ and $x = 2$.

Thus, the partial derivative is continuous except at $x = 2$ and $y = 0$.

Conclusion:

• The partial derivative of $F(x, y)$ with respect to $y$ is: $\frac{\partial F}{\partial y} = \frac{y(x - 2)}{\sqrt{y^2(x - 2)}}$
• The partial derivative is not continuous at $x = 2$ and $y = 0$, as there is a discontinuity at these points. For all other values of $x$ and $y$, the partial derivative is continuous.

Let me know if you'd like more details or have any further questions!

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Related Questions:

1. How do you find the partial derivative of a function involving absolute values?
2. What is the condition for the continuity of partial derivatives of a function?
3. Can you apply the chain rule to functions involving square roots and absolute values?
4. How does the square root function behave for negative values of the argument?
5. What is the significance of the domain when computing partial derivatives?

Tip:

When taking partial derivatives, always check the domain of the function to ensure that there are no undefined points (like division by zero or taking square roots of negative numbers).