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 follow these steps:

### Step 1: Differentiate the function
We start by differentiating \(y\):
\[ y = x^{\sin x} + \sqrt{x} \]

Using the sum rule of differentiation, we have:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^{\sin x}) + \frac{d}{dx}(\sqrt{x}) \]

### Step 2: Differentiate \(x^{\sin x}\)
To differentiate \(x^{\sin x}\), we will use logarithmic differentiation:
Let \(y_1 = x^{\sin x}\). Taking the natural logarithm on both sides:
\[ \ln y_1 = \sin x \cdot \ln x \]

Now, differentiate both sides with respect to \(x\):
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \]

Multiplying through by \(y_1\) gives:
\[ \frac{dy_1}{dx} = y_1 \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right) \]

Substituting back \(y_1 = x^{\sin x}\):
\[ \frac{dy_1}{dx} = x^{\sin x} \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right) \]

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

### Step 4: Combine the derivatives
Now we can combine the derivatives:
\[ \frac{dy}{dx} = x^{\sin x} \left(\cos x \cdot \ln x + \frac{\sin x}{x}\right) + \frac{1}{2\sqrt{x}} \]

### Step 5: Evaluate at \(x = \frac{\pi}{2}\)
Now we substitute \(x = \frac{\pi}{2}\) into the derivative:
1. Calculate \(x^{\sin x}\) at \(x = \frac{\pi}{2}\):
\[ \left(\frac{\pi}{2}\right)^{\sin\left(\frac{\pi}{2}\right)} = \left(\frac{\pi}{2}\right)^{1} = \frac{\pi}{2} \]

2. Calculate \(\cos\left(\frac{\pi}{2}\right) = 0\) and \(\sin\left(\frac{\pi}{2}\right) = 1\):
\[ \frac{\sin\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi} \]

3. Substitute into the derivative:
\[ \frac{dy}{dx} = \frac{\pi}{2} \left(0 \cdot \ln\left(\frac{\pi}{2}\right) + \frac{2}{\pi}\right) + \frac{1}{2\sqrt{\frac{\pi}{2}}} \]
Simplifying this:
\[ = \frac{\pi}{2} \cdot \frac{2}{\pi} + \frac{1}{2\sqrt{\frac{\pi}{2}}} \]
\[ = 1 + \frac{1}{2\sqrt{\frac{\pi}{2}}} \]
\[ = 1 + \frac{1}{\sqrt{2\pi}} \]

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