What is the 3rd Derivative of Cos(x) using this Derivative Formula? There is a general formula for the derivative of a function:
$$\frac{d^n}{dx^n}f(x)=\lim_{\epsilon\to0}\frac{1}{\epsilon^n}\sum_{j=0}^n{((-1)^j\frac{\Gamma(n+1)}{j!\Gamma{(n+1-j)}}f(x-j\epsilon))}$$
Where $\Gamma(x) $ is the Gamma function
I tried using the formula to evaluate the 3rd derivative of $\cos(x)$, but I get confused quickly. It would be very appreciated if someone could show a step by step solution to this problem. 
I'm totally aware the answer is $\sin(x)$, but what's the process to get to that solution?

We want
$$ \lim_{h \to 0} \frac{\cos{(x+3h)}-3\cos{(x+2h)+3\cos{(x+h)}-\cos{x}}}{h^3}. $$
Then
$$ \cos{(x+3h)}-\cos{(x+2h)} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} \\
-\cos{(x+2h)}+\cos{(x+h)} = 4\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} \\
\cos{(x+h)}-\cos{x} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)},
 $$
then
$$ -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} + 2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} = -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} \\
2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)}, $$
and finally
$$ -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} + \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^3\sin{\left( x + \frac{3h}{2} \right)}. $$
Then
$$ \lim_{h \to 0} \frac{1}{h^3}\left( 2\sin{\left(\frac{h}{2}\right)}\right)^3 = 1, $$
and the other term tends to $\sin{x}$, so the whole thing converges to $\sin{x}$. Exactly the same argument works for any number of derivatives, as can be shown by induction.
