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

Answer

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

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

### Step 2: Differentiate using the sum rule
Using the sum rule of differentiation, we have:
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
where \( u = x^x \) and \( v = (\sin x)^x \).

### Step 3: Differentiate \( u = x^x \)
To differentiate \( u = x^x \), we can use logarithmic differentiation:
1. Take the natural logarithm of both sides:
\[ \ln u = \ln(x^x) = x \ln x \]
2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \ln x) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]
3. Multiply both sides by \( u \) to isolate \( \frac{du}{dx} \):
\[ \frac{du}{dx} = u(\ln x + 1) = x^x (\ln x + 1) \]

### Step 4: Differentiate \( v = (\sin x)^x \)
Similarly, for \( v = (\sin x)^x \):
1. Take the natural logarithm:
\[ \ln v = \ln((\sin x)^x) = x \ln(\sin x) \]
2. Differentiate both sides:
\[ \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(x \ln(\sin x)) \]
Using the product rule again:
\[ \frac{d}{dx}(x \ln(\sin x)) = 1 \cdot \ln(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \ln(\sin x) + x \cot x \]
3. Multiply both sides by \( v \):
\[ \frac{dv}{dx} = v(\ln(\sin x) + x \cot x) = (\sin x)^x (\ln(\sin x) + x \cot x) \]

### Step 5: Combine the derivatives
Now, substituting back into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
\[ \frac{dy}{dx} = x^x (\ln x + 1) + (\sin x)^x (\ln(\sin x) + x \cot x) \]

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

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