While I can easily imagine the second derivative conveying the concavity, and the first derivative conveying the slope of any function in a graph. How do I visually understand the meaning of higher derivatives apart from the fact that they represent the rate of $(n-1)^{th}$ derivatives.

  • 1
    $\begingroup$ As a partial answer: When the first $n$ derivatives equal zero, the $(n+1)^{th}$ derivative tells you about increasing/decreasingness or convexity/concavity, so there's a little bit of that interpretation remaining (but with the caveat that these are overpowered by any non-zero lower derivative). $\endgroup$ Commented Aug 19, 2018 at 1:18
  • $\begingroup$ See this question for the third derivative: math.stackexchange.com/questions/14841/… $\endgroup$
    – present
    Commented Aug 19, 2018 at 1:50

1 Answer 1


I don't understand the relation, if you take $x \mapsto x^3$, you have :

$$ \left. \frac{d^2}{dx^2}x^3\right|_{x=0} = 0 $$

But :

$$ \frac{d^3}{dx^3}x^3 = 6 >0 $$

and $x \mapsto x^3$ have no maxima.

  • $\begingroup$ That's a good point, I think I might have made a mistake with the relation. Will correct it. $\endgroup$ Commented Aug 19, 2018 at 0:44
  • $\begingroup$ I have edited the question, please have a look. $\endgroup$ Commented Aug 19, 2018 at 0:46

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