### Applications of differentiation: Higher derivatives

### Higher derivatives

Let #f# be a function on an interval #I# and let #c# be a point of #I#. Suppose #f# is differentiable. Since #f'(x)# is also a function, we can examine its derivatives.

Higher derivative

Assume that #f# is differentiable. We can also consider the differentiability of #f'#. If the derivative of #f'# exists, then #f# is called **twice differentiable** on #I#. The notation of the derivative of #f'# is #f''#. For the derivative of #f'# in #c# we write #f''(c)# or #f^{(2)}(c)#. Instead of #f^{(2)}(x)# we also write #\dfrac{{\dd}^2}{{\dd}x^2}f(x)#.

If we continue along this line, we call, for each natural number #n\ge1#, the function #n#-**fold differentable** if #f#, #f'=f^{(1)}#, #f'' = f^{(2)}, \ldots, f^{(n-1)}# are differentiable. We write #f^{(0)} = f# and #f^{(n)}# for the derivative of #f^{(n-1)}#.

Furthermore, it is customary to write #\dfrac{{\dd}^n}{{\dd}x^n}f(x)# for #f^{(n)}(x)#.

Because #f'(x)# tells a lot about the shape of the graph of #f(x)#, we can similarly use #f''(x)# to learn more about #f'(x)# and #f(x)#. To this end we introduce the following concepts.

- #f# is called
**concave**(concave down) if the tangent lines to the graph of #f# are all above the graph. - #f# is called
**convex**(concave up) if the tangent lines of #f# are all below the graph. - #p\in I# is called an
**inflection point**if #p# is a point where the tangent line of the graph crosses the graph.

An inflection point is the transition between concave and convex, where the curve changes from concave to convex or the other way around.

See the following graph of a function #f# given by the blue curve.

The middle (grey) point is the only inflection point of #f# on the shown interval. Left of the inflection point, the (grey) tangent lies above the graph in the inflection point. Right of the inflection point, the (grey) tangent line lies below the graph of #f# in the inflection point.

Left of the inflection point, #f# is concave, because the (red) tangent of a random point left of the inflection point lies above the graph. Right of the inflection point, #f# is convex, because the (green) tangent of a random point right of the inflection point lies below the graph. In the inflection point, the shape of the graph (with increasing #x#-values) is changing from concave to convex in this example.

Second derivative test

Assume that #f# is twice differentiable on #I#.

- If #f''(x)\gt 0# for all #x\in I#, then #f# is convex on #I#.
- If #f''(x)\lt 0# for all #x\in I#, then #f# is concave on #I#.
- If #p# lies within #I# and is an inflection point of #f#, then #f''(p)=0#.
- If #p# lies within #I# and satisfies #f'(p)=0# and #f''(p)\lt0#, then #f# has a local maximum in #p#.
- If #p# lies within #I# and satisfies #f'(p)=0# and #f''(p)\gt0#, then #f# has a local minimum in #p#.

Since #f(x)# is convex, we have #f''(x) \gt 0#. This implies that #f'(x)# is increasing. So the slopes of the tangent line to #f# are increasing. As a consequence, the velocity of the function #f# is increasing.

Similarly, if #f(x)# is concave, then #f''(x) \lt 0#. This implies that #f'(x)# is decreasing. So the slopes of the tangent lines to #f# are decreasing. As a consequence, the velocity of the function #f# is decreasing.

In an inflection point we gave #f''(p)=0#. This means that #p# is a stationary point of #f'#. So before the point #p# the function #f(x)# has a decreasing velocity and form the point #p# it has an increasing velocity, or the other way around.

If #p# lies within #I# and satisfies #f'(p)=0# and #f''(p)=0#, we need some further research to see if #f# has a local maximum, a local minimum or none of both in #p#.

#f''(x)=# #336\cdot x^5+48\cdot x#

#f'''(x)=# #1680\cdot x^4+48#

After all, the

*power rule for differentiation*states: #\dfrac{\dd}{\dd x} \left(x^n\right)=n \cdot x^{n-1}# and the

*extended sum rule*states #\dfrac{\dd}{\dd x} \left(a \cdot f(x) +b \cdot g(x)\right)=a \cdot f'(x) = b \cdot g'(x)#. This gives:

\[\begin{array}{rcl}f'(x)&=&

\frac{\dd}{\dd x}\left(8\cdot x^7+8\cdot x^3+5\cdot x+5\right)\\

&=&8\cdot\frac{\dd}{\dd x}\left(x^{7}\right)+8\cdot\frac{\dd}{\dd x}\left( x^{3}\right)+5 \frac{\dd}{\dd x}\left( x\right)+\frac{\dd}{\dd x}(5)\\

&=&8\cdot 7\cdot x^{7-1}+8\cdot 3\cdot x^{3-1}+5+0\\

&=&56\cdot x^6+24\cdot x^2+5\end{array}\]

Now we apply the same rules to #f'(x)#:

\[f''(x)=336\cdot x^5+48\cdot x\tiny.\]

Finally, we apply the same rules to # f''(x)#:

\[f'''(x)=1680\cdot x^4+48\tiny.\]

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