Derivative of `(sinx)^(x)+sin^(-1)sqrtx` with respect to x is

Options

A. `(xcotx+logsinx)+(1)/(2sqrt(x-x^(2)))`
B. `(xcotx+logsinx)+(1)/(sqrt(x-x^(2)))`
C. `(sinx)(xcotx+log)+(1)/(sqrt(x-x^(2)))`
D. `(sinx)^(x)(xcotx+logsinx)+(1)/(2sqrt(x-x^(2)))`

Answer

To find the derivative of the function \( y = (\sin x)^x + \sin^{-1}(\sqrt{x}) \) with respect to \( x \), we will differentiate each term separately and then combine the results.

### Step 1: Differentiate \( y = (\sin x)^x \)

Let \( u = (\sin x)^x \). To differentiate this, we will use logarithmic differentiation.

1. Take the natural logarithm of both sides:
\[ \ln u = x \ln(\sin x) \]

2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \ln(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x \]
Here, we used the product rule for differentiation on \( x \ln(\sin x) \).

3. Rearranging gives:
\[ \frac{du}{dx} = u \left( \ln(\sin x) + x \cdot \frac{\cos x}{\sin x} \right) \]

4. Substitute back \( u = (\sin x)^x \):
\[ \frac{du}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) \]

### Step 2: Differentiate \( v = \sin^{-1}(\sqrt{x}) \)

Let \( v = \sin^{-1}(\sqrt{x}) \). We will use the chain rule here.

1. The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \). Therefore:
\[ \frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) \]

2. The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \):
\[ \frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(1 - x)}} \]

### Step 3: Combine the derivatives

Now we can combine the derivatives from both parts:
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substituting the values 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 \( y = (\sin x)^x + \sin^{-1}(\sqrt{x}) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) + \frac{1}{2\sqrt{x(1 - x)}} \]