Taylor Polynomial of $\arctan$ of given Degree and Error 
Replace the following function by its taylor polynomial of the given
  grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$

My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$$
f'(x) = \frac{1}{1+x^2} \\
f''(x) = \frac{2x}{(x^2+1)^2} \\
f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\
f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4}
$$
And by definition we know that
$$
T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k
$$
we get
$$
T_3(f,x,0) = x - \frac{x^3}{3}
$$
We know that 
$$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\
\le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4}
$$
$$
\Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\
\Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4}
$$
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).

This should be considered a minor addition to the answer by Mike Spivey. 
Note that the fourth derivative is $0$ at $0$. So the Taylor polynomial that we get by expanding until the third derivative is exactly the same as the Taylor polynomial we get by expanding until the fourth derivative!  
Thus you can use the formula for the remainder that involves the fifth derivative. There is not a lot of gain for the extra work, but there is some. The fifth derivative, I think, is $\dfrac{24(5x^4-10x^2+1)}{(x^2+1)^5}$. 
The analysis leads to an upper bound on the error which is about one-fifth of the error upper bound we get by using the Lagrange estimate without noting that stopping at $3$ gives the same polynomial as stopping at $4$.
Remark: You may have worked too hard in doing the arithmetic. Let's do exactly as you did, using the fourth derivative instead of the better fifth. 
At a certain stage, you reached
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right|
$$
Note that $c^2+1\gt 1$ and $|c^2-1|\lt 1$. So our fourth derivative has absolute value $\lt 24\cdot \dfrac{1}{10}$. We have harly given away anything in making this simple estimate. Now divide by $24$, multiply by $\left(\dfrac{1}{10}\right)^4$. We find that the absolute value of the error is $\lt 10^{-5}$. 
