Differentiate `log(x^(x)+cosec^(2)x)` w.r.t. x.

Answer

To differentiate the function \( y = \log(x^x + \csc^2 x) \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate the logarithmic function
We start with the function:
\[ y = \log(x^x + \csc^2 x) \]
Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{x^x + \csc^2 x} \cdot \frac{d}{dx}(x^x + \csc^2 x) \]

### Step 2: Differentiate \( x^x \)
To differentiate \( x^x \), we can use the property of logarithms. Let \( z = x^x \). Taking the natural logarithm of both sides gives:
\[ \log z = x \log x \]
Differentiating both sides with respect to \( x \) using implicit differentiation:
\[ \frac{1}{z} \frac{dz}{dx} = \log x + 1 \]
Thus, we have:
\[ \frac{dz}{dx} = z(\log x + 1) = x^x(\log x + 1) \]

### Step 3: Differentiate \( \csc^2 x \)
Next, we differentiate \( \csc^2 x \):
\[ \frac{d}{dx}(\csc^2 x) = 2 \csc^2 x \cdot (-\cot x) = -2 \csc^2 x \cot x \]

### Step 4: Combine the derivatives
Now we can substitute back into our derivative from Step 1:
\[ \frac{d}{dx}(x^x + \csc^2 x) = x^x(\log x + 1) - 2 \csc^2 x \cot x \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{x^x + \csc^2 x} \cdot \left( x^x(\log x + 1) - 2 \csc^2 x \cot x \right) \]

### Final Answer
Putting it all together, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{x^x(\log x + 1) - 2 \csc^2 x \cot x}{x^x + \csc^2 x} \]