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

Answer

To differentiate the function \( y = (\sin x)^x + \sin^{-1}(\sqrt{x}) \) with respect to \( x \), we will follow these steps:

### Step 1: Define the function
Let
\[ y = (\sin x)^x + \sin^{-1}(\sqrt{x}) \]

### Step 2: Differentiate using the sum rule
Using the sum rule of differentiation, we have:
\[ \frac{dy}{dx} = \frac{d}{dx}((\sin x)^x) + \frac{d}{dx}(\sin^{-1}(\sqrt{x})) \]

### Step 3: Differentiate \( u = (\sin x)^x \)
To differentiate \( u = (\sin x)^x \), we can use logarithmic differentiation. Taking the natural logarithm of both sides:
\[ \ln u = x \ln(\sin x) \]
Now differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x} \]
This simplifies to:
\[ \frac{du}{dx} = u \left( \ln(\sin x) + x \cot x \right) \]
Substituting back \( u = (\sin x)^x \):
\[ \frac{du}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) \]

### Step 4: Differentiate \( v = \sin^{-1}(\sqrt{x}) \)
Using the chain rule, we differentiate \( v = \sin^{-1}(\sqrt{x}) \):
\[ \frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1 - x)}} \]

### Step 5: Combine the derivatives
Now we can combine the derivatives:
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) + \frac{1}{2\sqrt{x(1 - x)}} \]

### Final Answer
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) + \frac{1}{2\sqrt{x(1 - x)}} \]

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