7. Higher-order derivatives

Part 1.: Exercises, solutions.
Part 2.: Exercises, solutions.

Higher-order derivatives

Obviously, if the derivative of a function is differentiable, we can calculate its derivative, the derivative of the derivative, f''(x), which describes how the first derivative changes. And further we can calculate the third, fourth, etc. derivatives. Generally n-th derivative will be denoted by f^{(n)} and f^{(n+1)}(x)=f^{(n)}(x).

E.g., for f(x)=x^3, we get:





and for any n\geq 4,


Condition for the existence of a local extremum

The following theorem holds. If for some n>0:

  • f is differentiable in x up to at least 2n-th derivative,
  • f'(x)=f''(x)=\ldots=f^{(2n-1)}=0,
  • f^{(2n)}(x)>0 (respectively: f^{(2n)}(x)<0),

then f has a minimum (respectively, maximum) in x.

E.g. f(x)=x^8-x^4 has maximum in x=0, because:

    \[f'(x)=8x^7-4x^3\Rightarrow f'(0)=0\]

    \[f''(x)=56x^6-12x^2\Rightarrow f''(0)=0\]

    \[f'''(x)=336x^5-24x\Rightarrow f'''(0)=0\]

    \[f^{(4)}(x)=1680x^4-24\Rightarrow f^{(4)}(0)=-24\]

Convex and concave functions, inflection points

We shall say that f is convex on an interval (a,b), if its graph between any two points x,y\in (a,b) is under its secant line given by those points. In other words if for any x,y\in(a,b) and t\in [0,1]:


If the reverse inequality holds, the function f is concave. If f is convex on (a,b) and concave on (b,c) (or reversely) and continuous in c, we will say that c is an inflection point.

The following theorems hold:

  • if f' exists on a given interval and is increasing (respectively, decreasing), then f is convex (respectively, concave) on this interval,
  • if f'' exists on a given interval and is always positive (respectively, negative), then f f is convex (respectively, concave) on this interval,

E.g.: f(x)=x^3. We get f'(x)=3x^2, f''(x)=6x is positive on (0,\infty) so f is convex on this interval and negative on (-\infty,0) — so f is concave on it. Point 0 is an inflection point.