If $D$ be the differentiation operator on $V$. Find $D^*$. 
Let $V$ be the vector space of the polynomials over $R$ of degree less than or
  equal to $3$ with the inner product space $(f|g)=\int_{0} ^{1}f(t)g(t) dt$, and let $D$ be the differentiation operator on V. Find $D^*$

Attempt
As I did the calculation was very large, so I will explain what I did step by step so that you say if I made a mistake or I hit
STEP 1: First I considered a canonical basis of $R _{\leq 2} [X]$ that is $ \{ 1,x,x^2,x^3\}$. I did the Gram-Schmidt Process to orthogonalize this base. Resulted in
$$ \left\{ 1, x-\frac{1}{2}, x^2-x+ \frac{1}{6}, x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right\}$$
STEP 2: 
Orthonormalize this base, dividing each element by its norm (in this part I may have been wrong because they gave many calculations). Resulted in
$$ \{ 1, \sqrt{12} \left(x-\frac{1}{2} \right) , \sqrt{180} \left( x^2-x+ \frac{1}{6} \right) , \sqrt{\frac{33600}{18772}} \left( x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right) \}$$
STEP 3: Write the matrix of $D$
$$\begin{bmatrix}0&\sqrt{12}&0& \frac{1}{10} \sqrt{\frac{33600}{18772}}\\0&0&2\sqrt{15}& 0\\ 0&0&0& \sqrt{\frac{33600}{33378960}} \\ 0&0&0& 0 \end{bmatrix}$$
STEP4: The matrix of $D^*$ is the conjugate transpose of the matrix of $D$,(in this case only transpose) according to the corollary: 

Let $V$ be a finite-dimensional inner product space, and let
  $T$ be a linear operator on $V$. In any orthonormal basis for $V$, the matrix of $T^*$ is the conjugate transpose of the matrix of $T$.

At least the idea is correct? Is there any easier way to do this exercise?
 A: Yes, the idea is correct (I didn't verify the coefficients in the orthonormalization).
Here's an alternative method: Integrating by parts gives
$$\langle f, Dg \rangle = \int_0^1 f g' dx = f(1) g(1) - f(0) g(0) + \int_0^1 - f' g \,dx = f(1) g(1) - f(0) g(0) + \langle -Df, g \rangle.$$
Thus, if we can find an operator $S$ such that $\langle S f, g \rangle = f(1) g(1) - f(0) g(0)$, we will have
$$\langle f, D g \rangle = \langle (S - D) f, g \rangle$$ and thus $$D^* = S - D .$$
Remark Up to this point, the formulation works for all functions in $L^2([0, 1])$, and if we enlarge our space to include the Dirac delta function, by definition we can write $S = \delta(t - 1) - \delta(t)$ and thus $D^* = -D + \delta(t - 1) - \delta(t) .$)
To find an explicit formula for $S$, we look for polynomials $r, s \in V$ such that
$$\langle S f , g \rangle = \langle f(1) r - f(0) s , g \rangle$$
for all $f , g \in V$. On the one hand, $\langle S f, g \rangle = \langle D f, g \rangle + \langle f, D g \rangle$, and except for $k = l = 0$, this gives $\langle S(x^k), x^l \rangle = 1$. So, writing $r = \sum_{i = 0}^3 r_i x^i$ and $s = \sum_{i = 0}^3 s_i x^i$ and integrating against monomials $f = x^k, g = x^l$ gives (for $k > 0$)
$$1 = \langle S (x^k), x^l \rangle = \langle r, x^l \rangle .$$
Writing this in terms of $r_0, r_1, r_2, r_3$ for $l = 0, 1, 2, 3$ and integrating gives rise to the system
$$\pmatrix{1&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}\\\frac{1}{2}&\frac{1}{3}&\frac{1}{4}&\frac{1}{5}\\\frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\frac{1}{6}\\\frac{1}{4}&\frac{1}{5}&\frac{1}{6}&\frac{1}{7}} \pmatrix{r_0\\r_1\\r_2\\r_3} = \pmatrix{1\\1\\1\\1} .$$ Solving gives
$$r_0 = -4, r_1 = 60 , r_2 = -180, r_3 = 140 ,$$
so $r(x) = 4 (35 x^3 - 45 x^2 + 15 x - 1)$.
A similar argument (and using that $\langle S(1), 1 \rangle = 0$) gives that $s(x) = -4 (35 x^3 - 60 x^2 + 30 x - 4)$. Putting everything together gives
$$\boxed{D^* f = -D f + 4 [f(1) (35 x^3 - 45 x^2 + 15 x - 1) - f(0) (35 x^3 - 60 x^2 + 30 x - 4)]} .$$
The matrix on the left-hand side is the $4 \times 4$ Hilbert matrix; there is a general formula for the inverse of the analogous $n \times n$ matrix, which means we can write down an explicit formula for the adjoint for the differentiation operator on the space of polynomials of degree $\leq n$ for general $n$.
A: Yes, your idea is correct, and this is the standard way to find the adjoint of an operator.
Another attempt could be using integration by parts:
\begin{align}
\langle f',g\rangle &= \int_0^1 f'(t)g(t)\,dt \\
&= f(t)g(t)\bigg|_0^1 - \int_0^1 f(t)g'(t)\,dt \\
&= f(1)g(1) - f(0)g(0) - \langle f,g'\rangle \\
\end{align}
Hence find an operator $A$ such that $\langle f,Ag\rangle =  f(1)g(1) - f(0)g(0)$ and then we will have $$\langle f',g\rangle = \langle f,(A-D)g\rangle$$
Therefore the adjoint is $D^* = A-D$.

To find $A$ explicitly, let's find a polynomial $p(x) = ax^3+bx^2+cx+d \in \mathbb{R}_{\le 3}[x]$ such that for all $f \in \mathbb{R}_{\le 3}[x]$ holds
$$\int_0^1 f(t)p(t)\,dt = f(0)$$
In particular, for $f(x) = 1,x,x^2,x^3$ we get
$$1 = \int_0^1 p(t)\,dt = \frac{a}4 + \frac{b}3 + \frac{c}2 + d$$
$$0 = \int_0^1 tp(t)\,dt = \frac{a}5 + \frac{b}4 + \frac{c}3 + \frac{d}2$$
$$0 = \int_0^1 t^2p(t)\,dt = \frac{a}6 + \frac{b}5 + \frac{c}4 + \frac{d}3$$
$$0 = \int_0^1 t^3p(t)\,dt = \frac{a}7 + \frac{b}6 + \frac{c}5 + \frac{d}4$$
Solving the system yields $p(x) = -140x^3+ 240x^2 -120x +16$.
Similarly we find that the polynomial $q(x) = 140x^3-180x^2 +60x -4$ satisfiels
$$\int_0^1 f(t)q(t)\,dt = f(1), \quad\forall f \in \mathbb{R}_{\le 3}[x]$$
Therefore, if we set $Ag := (p-q)g$, we get
$$\langle f,Ag\rangle = \int_0^1f(t)g(t)(p(t) - q(t))\,dt = f(1)g(1) - f(0)g(0)$$
