The value of `(dy)/(dx)` at `x=(pi)/(2)`, where y is given by
`y=x^(sinx)+sqrt(x)`, is

Options

A. `1+(1)/(sqrt(2pi))`
B. 1
C. `(1)/(2pi))`
D. `1-(1)/(sqrt(2pi))`

Answer

To find the value of \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \) for the function \( y = x^{\sin x} + \sqrt{x} \), we will differentiate the function and evaluate it at the specified point.

### Step-by-Step Solution:

1. **Differentiate the function**:
We have:
\[ y = x^{\sin x} + \sqrt{x} \]
We will differentiate this function with respect to \( x \).

For the first term \( x^{\sin x} \), we can use logarithmic differentiation:
\[ \ln y_1 = \sin x \ln x \]
Differentiating both sides:
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x} \]
Thus, we have:
\[ \frac{dy_1}{dx} = y_1 \left( \cos x \ln x + \frac{\sin x}{x} \right) \]
Substituting back for \( y_1 \):
\[ \frac{dy_1}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) \]

For the second term \( \sqrt{x} \):
\[ \frac{dy_2}{dx} = \frac{1}{2\sqrt{x}} \]

Therefore, the total derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + \frac{1}{2\sqrt{x}} \]

2. **Evaluate at \( x = \frac{\pi}{2} \)**:
Now we substitute \( x = \frac{\pi}{2} \):
\[ y_1 = \left( \frac{\pi}{2} \right)^{\sin\left(\frac{\pi}{2}\right)} = \left( \frac{\pi}{2} \right)^{1} = \frac{\pi}{2} \]
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus,
\[ \frac{dy_1}{dx} = \frac{\pi}{2} \left( 0 \cdot \ln\left(\frac{\pi}{2}\right) + \frac{1}{\frac{\pi}{2}} \right) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1 \]

For the second term:
\[ \frac{dy_2}{dx} = \frac{1}{2\sqrt{\frac{\pi}{2}}} = \frac{1}{2 \cdot \sqrt{\frac{\pi}{2}}} = \frac{1}{\sqrt{2\pi}} \]

3. **Combine the results**:
Now we can combine the derivatives:
\[ \frac{dy}{dx} = 1 + \frac{1}{\sqrt{2\pi}} \]

### Final Answer:
Thus, the value of \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \) is:
\[ \frac{dy}{dx} = 1 + \frac{1}{\sqrt{2\pi}} \]