Domain of $x^{\sin x+\cos x}$ Why domain of $x^{\sin x+ \cos x}$ is $x\ge 0$ while domain of $x^x$ is $x&gt;0$ 
And if we take the natural logarithim for
 $f(x)= x^{\sin x+ \cos x}$
we get $\ln(f(x))=(\sin x+ \cos x) \ln(x)$ and if $x=0$  we will get  $\ln(0)$ ?
For $f(x)=x^x$, $\text{dom}(f)=\{x | x\in \mathbb{Z}_{-}\} \cup \{x| x \in \mathbb{R}_{+}\}$. But $0 \notin \text{dom}(f)$. In the case of $x^{\sin(x)+\cos(x)}$, you can calculate the value at $x=0$, because $0^x = 0$ $\forall x &gt; 0$.
About the first expression: if you plug $x = 0$ in, you get $0^{0+1} = 0$. In the second you get $0^0$, which is indeterminate. That is why the second expression is not defined for $x=0$. 

You cannot compose $\ln$ with $x^{\sin x + \cos x}$ since this expression at $0$ returns $0$. The logarithm is not defined at $0$.