Differentiate the function $y = x^{x} - 2^{\sin x}$ with respect to $x$.

Let $y = x^{x} - 2^{\sin x}$.
Let $u = x^{x}$ and $v = 2^{\sin x}$.
Then $y = u - v$,so $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.
For $u = x^{x}$,taking logarithms on both sides: $\log u = x \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{du}{dx} = u(1 + \log x) = x^{x}(1 + \log x)$.
For $v = 2^{\sin x}$,taking logarithms on both sides: $\log v = \sin x \cdot \log 2$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = \log 2 \cdot \frac{d}{dx}(\sin x) = \log 2 \cdot \cos x$.
Thus,$\frac{dv}{dx} = v \cdot \cos x \cdot \log 2 = 2^{\sin x} \cos x \log 2$.
Substituting these back,we get $\frac{dy}{dx} = x^{x}(1 + \log x) - 2^{\sin x} \cos x \log 2$.