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It's well-known that $f'(x),f''(x)$ determine the monotonicity and the concavity of $f(x)$ respectively. Besides, we can see $f'(x)$ and $f''(x)$ from the graph of $f(x)$.

But how about $f'''(x)$ and other higher orderd derivative? What's its geometrical meaning on the graph of $f(x)$? Or, limited by our human-being's perception, we can't see high-orderd derivative?

WuKong
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2 Answers2

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Usually we don't give a specific geometrical meaning to $f'''(x)$ with reference to $f(x)$ but we can observe that $$f'''(x)=(f''(x))'=(f'(x))''$$

and therefore $f'''(x)$ represents

  • the slope of $f''(x)$
  • the convexity/concavity of $f'(x)$

With reference to $f(x)$, we can be interested in the higher order derivatives to determine the nature of critical points.

Notably, when derivatives exist if $f'(x_0)=0$ and $\exists k \geq 2$ s.t. $f^k(x_0) \neq 0$ then

  • when k is even we have a max/min in $x_0$ (depending on the sign)

  • when k is odd we have an inflection point in $x_0$

user
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    I can agree with you！But what I want to know is that where $f'''(x)$ is on the graph of $f(x)$ itself... – WuKong Jul 03 '18 at 12:53
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    Otherwise, you can always say that $f^{(n)}(x)$ is the slope of $f^{(n-1)}(x)$. That seems too trivial. – WuKong Jul 03 '18 at 12:56
  • @mengdie1982 I'm sorry that it is trivial but there is no other general meaning for that. – user Jul 03 '18 at 12:58
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If $f(t)$ represents position at time $t$, then the third derivative $f'''(t)$ is sometimes called jerk and represents change in acceleration at time $t$. See this link on Wikipedia for more information.

MPW
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  • I know that physical concept. You can feel the velocity when you are driving, and the acceleration when you are stepping on the gas. But how you feel the so-called JERK? – WuKong Jul 03 '18 at 13:03