Differentiate the following w.r.t. x :
`x^(x)sin^(-1)sqrtx`

Answer

To differentiate the function \( y = x^x \sin^{-1}(\sqrt{x}) \) with respect to \( x \), we will use the product rule and the chain rule. Let's break this down step by step.

### Step 1: Identify the components
We have:
- \( u = x^x \)
- \( v = \sin^{-1}(\sqrt{x}) \)

### Step 2: Differentiate using the product rule
The product rule states that if \( y = uv \), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step 3: Differentiate \( u = x^x \)
To differentiate \( u = x^x \), we can use logarithmic differentiation:
1. Take the natural logarithm:
\[ \ln(u) = x \ln(x) \]
2. Differentiate both sides:
\[ \frac{1}{u} \frac{du}{dx} = \ln(x) + 1 \]
3. Multiply by \( u \):
\[ \frac{du}{dx} = x^x (\ln(x) + 1) \]

### Step 4: Differentiate \( v = \sin^{-1}(\sqrt{x}) \)
Using the chain rule:
1. The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
2. Therefore, for \( v = \sin^{-1}(\sqrt{x}) \):
\[ \frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1-x)}} \]

### Step 5: Combine using the product rule
Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule:
\[ \frac{dy}{dx} = x^x \cdot \frac{1}{2\sqrt{x(1-x)}} + \sin^{-1}(\sqrt{x}) \cdot x^x (\ln(x) + 1) \]

### Final Result
Thus, the derivative of \( y = x^x \sin^{-1}(\sqrt{x}) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^x \cdot \frac{1}{2\sqrt{x(1-x)}} + x^x \sin^{-1}(\sqrt{x}) (\ln(x) + 1) \]