To solve the problem, we need to find the partial derivative $\frac{\partial f}{\partial x}$ at the point $(\frac{\pi}{2}, \frac{\pi}{2})$, given the function:

$f(x, y) = x + y \sin(x) \text{ and } x^2 + y^2 = 1$

Step 1: Compute the Partial Derivative $\frac{\partial f}{\partial x}$

The function $f(x, y) = x + y\sin(x)$ is dependent on $x$ and $y$. The partial derivative of $f$ with respect to $x$ is calculated as follows:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(x + y\sin(x)\right)$

Differentiate each term:

• The derivative of $x$ with respect to $x$ is $1$.
• The derivative of $y\sin(x)$ with respect to $x$ is $y\cos(x)$ because $y$ is treated as a constant.

So, we get:

$\frac{\partial f}{\partial x} = 1 + y\cos(x)$

Step 2: Evaluate at the Given Point $(\frac{\pi}{2}, \frac{\pi}{2})$

Substitute $x = \frac{\pi}{2}$ and $y = \frac{\pi}{2}$ into the partial derivative:

$\frac{\partial f}{\partial x}\bigg|_{\left(\frac{\pi}{2}, \frac{\pi}{2}\right)} = 1 + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right)$

Since $\cos\left(\frac{\pi}{2}\right) = 0$, this simplifies to:

$\frac{\partial f}{\partial x}\bigg|_{\left(\frac{\pi}{2}, \frac{\pi}{2}\right)} = 1 + \frac{\pi}{2} \cdot 0 = 1$

Conclusion

The correct answer is $\boxed{1}$.

Would you like further details or have any questions? Here are some related questions to explore:

1. How do you compute partial derivatives for functions involving multiple variables?
2. What are the geometric interpretations of partial derivatives?
3. How does the chain rule apply when differentiating implicit functions?
4. What is the significance of the point $(\frac{\pi}{2}, \frac{\pi}{2})$ on the unit circle?
5. How would the result change if the function $f(x, y)$ had a different form?

Tip: Always check trigonometric identities and values at special angles like $\frac{\pi}{2}$ or $\pi$ to simplify calculations.