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.
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1As 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). – Milo Brandt Aug 19 '18 at 01:18
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See this question for the third derivative: https://math.stackexchange.com/questions/14841/what-is-the-meaning-of-the-third-derivative-of-a-function-at-a-point – present Aug 19 '18 at 01:50
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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.
tmaths
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That's a good point, I think I might have made a mistake with the relation. Will correct it. – Utkarsh Verma Aug 19 '18 at 00:44
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