Let $V$ be the space of complex polynomials on $[0,1]$. Is the differentiation operator self-adjoint? 
Suppose $V$ is the space of all complex polynomials of degree $N$ or less restricted to $[0,1]$. Equipped with the inherited inner product $\langle f,g \rangle = \int_0^1 f(t) \overline{g(t)} dt$. 

So for the setup above, let $D$ be the differentiation operator so
$$
D(a_0 + a_1 t + \cdots + a_N t^N) = a_1 + 2 a_2 t + \cdots + N a_N t^{N-1}
$$
Is $D$ is self adjoint or skew adjoint?
Attempt/Thoughts
In thinking about it, I know I can write the inner product as
$$
\int_0^1 f \overline g dt = \sum_{i,j=0}^N \frac{a_i \overline b_j}{i+j+1}
$$
with $a_k$ and $b_k$ as the coefficients of polynomials $f$ and $g$, respectively. Then
$$
\langle Df, g \rangle = \sum_{i=1,j=0}^N \frac{i a_i \overline b_j}{i+j}
$$
and
$$
\langle f, Dg \rangle = \sum_{i=0,j=1}^N \frac{j a_i \overline b_j}{i+j}
$$
so it would seem that $D$ is neither self adjoint or skew adjoint. Is this the right way to approach my question?
Another thought I had was integration by parts, but that didn't seem produce anything useful when I tried it.
 A: As already established in comments, $D$ is neither self-adjoint nor skew adjoint. In my opinion, the reason for this behaviour lies in the fact that $D$ lowers the degree of the polynomial it is applied to.
Consider an orthonormal basis of $V$ consisting of polynomials of increasing degree: 
$$
\mathcal B = (P_0, P_1, \ldots, P_N).$$
By this I mean that the degree of $P_j$ is $j$ and
$$\int_0^1P_jP_k\,dx = \delta_{jk}.$$Such a basis is obtained by applying the Gram-Schmidt process to the usual basis $(1, x, x^2,\ldots x^N)$ (in the language of special functions, those are the normalized shifted Legendre polynomials). 
The operator $D$ is self adjoint (or skew adjoint) if and only if its associated matrix with respect to $\mathcal B$ is symmetric (or skew symmetric). Since 
$$
DP_0=0,\quad DP_k=\sum_{j=0}^{k-1}c_{kj}P_j, $$
this matrix has the form 
$$\begin{bmatrix} 
0 & * & * & \ldots & * \\ 
0 & 0 & * & \ldots & * \\
0 & 0 & 0 & \ldots & * \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & \ldots & * \\ 
0 & 0 & 0 & \ldots & 0 \\
\end{bmatrix}$$
and clearly it has neither of the required symmetry properties. 
