Find `(dy)/(dx)`, when:
`y=x^(x)-2^(sinx)`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = x^x - 2^{\sin x}\), we will differentiate each term separately.

### Step 1: Differentiate \(x^x\)
To differentiate \(x^x\), we can use logarithmic differentiation.

1. Let \(t = x^x\).
2. Taking the natural logarithm of both sides:
\[ \ln t = x \ln x \]
3. Now differentiate both sides with respect to \(x\):
\[ \frac{1}{t} \frac{dt}{dx} = \ln x + 1 \]
4. Rearranging gives:
\[ \frac{dt}{dx} = t(\ln x + 1) \]
5. Substituting back for \(t\):
\[ \frac{dt}{dx} = x^x(\ln x + 1) \]

### Step 2: Differentiate \(2^{\sin x}\)
To differentiate \(2^{\sin x}\), we can use the chain rule:

1. The derivative of \(a^{u}\) is given by \(a^u \ln a \cdot \frac{du}{dx}\).
2. Here, \(a = 2\) and \(u = \sin x\):
\[ \frac{d}{dx}(2^{\sin x}) = 2^{\sin x} \ln 2 \cdot \cos x \]

### Step 3: Combine the results
Now we can combine the derivatives of both terms to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^x) - \frac{d}{dx}(2^{\sin x}) \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = x^x(\ln x + 1) - 2^{\sin x} \ln 2 \cdot \cos x \]

### Final Answer
Thus, the final result is:
\[ \frac{dy}{dx} = x^x(\ln x + 1) - 2^{\sin x} \ln 2 \cdot \cos x \]