Differentiate the following w.r.t. x :
`(sinx)^(x)`

Answer

To differentiate the function \( y = (\sin x)^x \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process.

\[ \ln y = \ln((\sin x)^x) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the equation as:

\[ \ln y = x \ln(\sin x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). Remember to use implicit differentiation on the left side.

\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln(\sin x)) \]

Using the chain rule on the left side:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln(\sin x)) \]

### Step 4: Differentiate the right side using the product rule
Now we apply the product rule on the right side. Let \( u = x \) and \( v = \ln(\sin x) \):

\[ \frac{d}{dx}(x \ln(\sin x)) = \frac{du}{dx} v + u \frac{dv}{dx} \]

Calculating each derivative:

- \( \frac{du}{dx} = 1 \)
- \( v = \ln(\sin x) \) and \( \frac{dv}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \)

Putting it all together:

\[ \frac{d}{dx}(x \ln(\sin x)) = 1 \cdot \ln(\sin x) + x \cdot \cot x \]

### Step 5: Substitute back into the equation
Now we substitute back into our equation:

\[ \frac{1}{y} \frac{dy}{dx} = \ln(\sin x) + x \cot x \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \) gives:

\[ \frac{dy}{dx} = y \left( \ln(\sin x) + x \cot x \right) \]

### Step 7: Substitute \( y \) back in
Since \( y = (\sin x)^x \), we substitute back:

\[ \frac{dy}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) \]

### Final Answer
Thus, the derivative of \( y = (\sin x)^x \) with respect to \( x \) is:

\[ \frac{dy}{dx} = (\sin x)^x \left( \ln(\sin x) + x \cot x \right) \]

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