There is a general formula for the derivative of a function:


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?

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    $\begingroup$ Recommended tip: Don't use that definition. Just take the first derivative 3 times. $\endgroup$ Commented Oct 16, 2017 at 18:36
  • $\begingroup$ I know of that method, but I’m looking for an answer utilizing the formula I stated. $\endgroup$
    – Tom Himler
    Commented Oct 16, 2017 at 18:39
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    $\begingroup$ "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. $\endgroup$ Commented Oct 16, 2017 at 18:40
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    $\begingroup$ Good exericse to warm up the brain] $\endgroup$
    – Guy Fsone
    Commented Oct 16, 2017 at 18:47
  • $\begingroup$ That formula cries desperately for refactoring. $\endgroup$
    – Thom Smith
    Commented Oct 16, 2017 at 22:02

2 Answers 2


$\Gamma(n+1)$ 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 $\ddot\frown$.

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    $\begingroup$ Does this formula for the n-th derivative have a name or does anyone know it's originator ? Just curious. $\endgroup$ Commented Oct 17, 2017 at 1:34
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    $\begingroup$ To answer my own question I believe it's the Grünwald–Letnikov derivative although please correct me if someone knows that's wrong. $\endgroup$ Commented Oct 17, 2017 at 4:13

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.


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