# 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?

• Recommended tip: Don't use that definition. Just take the first derivative 3 times. Commented Oct 16, 2017 at 18:36
• I know of that method, but I’m looking for an answer utilizing the formula I stated. Commented Oct 16, 2017 at 18:39
• "I tried using the formula... but I get confused quickly." In order to receive the best possible answers, it would be helpful to know what you can and can't do, so please include your own efforts. Commented Oct 16, 2017 at 18:40
• Good exericse to warm up the brain] Commented Oct 16, 2017 at 18:47
• That formula cries desperately for refactoring. Commented Oct 16, 2017 at 22:02

$\Gamma(n+1)$ is an affected way of writing $n!$. So the RHS is $$\newcommand{\ep}{\epsilon}\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\frac{n!}{j!(n-j)!}f(x+j\ep)= \lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\binom{n}{j}f(x+j\ep).$$ This isn't quite correct: there's a sign error, it should be $$(-1)^n\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\binom{n}{j}f(x+j\ep) =\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^{n-j}\binom{n}{j}f(x+j\ep).$$ For $n=3$ it should be $$\lim_{\ep\to0}\frac{f(x+3\ep)-3f(x+2\ep)+3f(x+\ep)-f(x)}{\ep^3}.$$
When in addition $f(x)=\cos(x)$ then we get $$f(x+\ep)=\cos \ep\cos x-\sin \ep\sin x,$$ $$f(x+2\ep)=(\cos^2 \ep-\sin^2\ep)\cos x-2\sin\ep\cos\ep\sin x,$$ etc. You eventually get $$\lim_{\ep\to 0}\frac{G(\ep)\cos x+H(\ep)\sin x}{\ep^3}$$ where $G(\ep)/\ep^3$ and $H(\ep)/\ep^3$ are functions that should tend to $0$ and $1$ as $\ep\to0$. I don't want to go into the details though $\ddot\frown$.
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.