If ` y=x^(sin x)+ sqrt(x) ,then " at "=(pi)/( 2), (dy)/(dx)=`

Options

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

Answer

To solve the problem, we need to find the derivative of the function \( y = x^{\sin x} + \sqrt{x} \) and then evaluate it at \( x = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Differentiate \( y \)**:
We have:
\[ y = x^{\sin x} + \sqrt{x} \]
We need to find \( \frac{dy}{dx} \).

2. **Differentiate \( x^{\sin x} \)**:
To differentiate \( x^{\sin x} \), we can use logarithmic differentiation. Let:
\[ z = x^{\sin x} \]
Taking the natural logarithm of both sides:
\[ \ln z = \sin x \cdot \ln x \]
Now differentiate both sides with respect to \( x \):
\[ \frac{1}{z} \frac{dz}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]
Rearranging gives:
\[ \frac{dz}{dx} = z \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]
Substituting back \( z = x^{\sin x} \):
\[ \frac{dz}{dx} = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) \]

3. **Differentiate \( \sqrt{x} \)**:
The derivative of \( \sqrt{x} \) is:
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

4. **Combine the derivatives**:
Now, we can combine the derivatives to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dz}{dx} + \frac{d}{dx}(\sqrt{x}) = x^{\sin x} \left( \cos x \cdot \ln x + \frac{\sin x}{x} \right) + \frac{1}{2\sqrt{x}} \]

5. **Evaluate \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \)**:
Now we substitute \( x = \frac{\pi}{2} \):
- First, calculate \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \).
- Therefore:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = \left(\frac{\pi}{2}\right)^{1} \left(0 \cdot \ln\left(\frac{\pi}{2}\right) + \frac{1}{\frac{\pi}{2}}\right) + \frac{1}{2\sqrt{\frac{\pi}{2}}} \]
Simplifying this:
\[ = \frac{\pi}{2} \cdot \frac{2}{\pi} + \frac{1}{2\sqrt{\frac{\pi}{2}}} \]
\[ = 1 + \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{\pi}{2}}} = 1 + \frac{1}{\sqrt{2\pi}} \]

6. **Final Answer**:
Therefore, the value of \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \) is:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{2}} = 1 + \frac{1}{\sqrt{2\pi}} \]