If $X$ is a smooth variety, how do we see that $c_1(T_X^*)$ is the canonical divisor? I can see this for the projective space: if $X=\mathbb{P}^n$, then the Euler exact sequence $$ 0\to O\to O(1)^{n+1}\to T_X\to 0 $$
gives $c_1(T_X)=(n+1)[H]$, and we know $\omega_X=O(-n-1)$. But how do we prove this in general?
In general, if $E$ is any vector bundle on $X$ (say, of rank $r$), one can talk about the Chern roots of $E$. These are defined to be the classes $\alpha_1,\dots,\alpha_r$ such that che Chern polynomial $$c_t(E)=\sum_{i=0}^rc_i(E)t^i\in A^\ast(X)[t]$$ can be written as $$c_t(E)=\prod_{i=1}^r(1+\alpha_it).$$ For instance, if $E$ has a filtration with line bundle quotients $L_1,\dots,L_r$, then $\alpha_i$ can be regarded as the Chern class $c_1(L_i)$ for $i=1,\dots,r$ (by repeatedly applying Whitney's formula). This allows to identify: $$c_i(E)=s_i(\alpha_1,\dots,\alpha_r),$$ the $i$-th symmetric function in the $\alpha_i$'s. For instance, $$c_1(E)=\alpha_1+\dots+\alpha_r.$$
For $1\leq p\leq r$, the Chern roots of $\wedge^pE$ are the symmetric functions $$\alpha_{j_1}+\dots+\alpha_{j_p},\qquad 1\leq j_1<\dots<j_p\leq r.$$
Now, the exterior power $\wedge^rE$ is a line bundle, so the unique Chern root is the sum $\alpha_1+\dots+\alpha_r$, but this equals $c_1(E)$. Hence in general: $$\fbox{$c_1(E)=c_1(\wedge^rE)$.}$$
In your case, $E=T^\ast_X$, and $\omega_X=\wedge^{\dim X}E$.