Exercises: Derivatives 2 – "Math for Non-Geeks"

Derivatives of higher order

Exercise 2

Exercise (Exactly one/two/three times differentiable functions)

Provide an example of a

1. function $f\in C^{0}(\mathbb {R} )\setminus C^{1}(\mathbb {R} )$
2. function that is differentiable, but not continuously differentiable on $\mathbb {R}$
3. function $f\in C^{2}(\mathbb {R} )\setminus C^{3}(\mathbb {R} )$

Solution (Exactly one/two/three times differentiable functions)

Solution sub-exercise 1:

$f(x)=|x|$ or $f(x)={\begin{cases}x\sin \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$ or $f(x)={\begin{cases}x\cos \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$

Solution sub-exercise 2:

$f(x)={\begin{cases}x^{2}\sin \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$ or $f(x)={\begin{cases}x^{2}\cos \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$

Solution sub-exercise 3:

$f(x)=x\cdot |x|$ or $f(x)={\begin{cases}x^{3}\sin \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$ or $f(x)={\begin{cases}x^{3}\cos \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$

Solution sub-exercise 4:

$f(x)=x^{2}\cdot |x|$ or $f(x)={\begin{cases}x^{5}\sin \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$ or $f(x)={\begin{cases}x^{5}\cos \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$

Hint

We can successively proceed with the construction of functions and obtain some $f\in C^{k}(\mathbb {R} )\setminus C^{k+1}(\mathbb {R} )$ for all $k\in \mathbb {N}$, with

$f(x)=x^{k}\cdot |x|$ or $f(x)={\begin{cases}x^{2k+1}\sin \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$ or $f(x)={\begin{cases}x^{2k+1}\cos \left({\frac {1}{x}}\right)&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0.\end{cases}}$

Exercise 3

Exercise (Application of the Leibniz rule)

Determine the following derivatives using the Leibniz rule

1. $(x\sin(x))^{(10)}$
2. $(x^{3}\ln(x))^{(2016)}$ for $x>0$

Solution (Application of the Leibniz rule)

Solution sub-exercise 1:

The functions ${\text{id}}$ and $\sin$ are arbitrarily often differentiable on $\mathbb {R}$ . Hence there is

${\begin{aligned}(x\sin(x))^{(10)}&{\underset {\text{rule}}{\overset {\text{Leibniz-}}{=}}}\sum _{k=0}^{10}{\binom {10}{k}}(x)^{(k)}(\sin(x))^{(10-k)}\\[0.3em]&\color {Gray}\left\downarrow \ (x)'=1,\ (x)^{(k)}=0{\text{ for }}k\geq 2,\ (\sin(x))^{(4k+1)}=\cos(x){\text{ and }}(\sin(x))^{(4k+2)}=-\sin(x){\text{ for all }}k\in \mathbb {N} \right.\\[0.3em]&={\binom {10}{0}}(x)^{(0)}(\sin(x))^{(10)}+{\binom {10}{1}}(x)'(\sin(x))^{(9)}\\[0.3em]&={\binom {10}{0}}x(-\sin(x))+{\binom {10}{1}}\cos(x)\\[0.3em]&=-x\sin(x)+10\cos(x)\end{aligned}}$

Solution sub-exercise 2:

The functions $x\mapsto x^{3}$ and $\ln$ are arbitrarily often differentiable on $\mathbb {R} ^{+}$ . Hence there is

${\begin{aligned}(x^{3}\ln(x))^{(2016)}&{\underset {\text{rule}}{\overset {\text{Leibniz-}}{=}}}\sum _{k=0}^{2016}{\binom {2016}{k}}(x)^{(k)}(\ln(x))^{(2016-k)}\\[0.3em]&\color {Gray}\left\downarrow \ (x^{3})'=3x^{2},\ (x^{3})''=6x,\ (x^{3})'''=6,\ (x)^{(k)}=0{\text{ for }}k\geq 4,\ (\ln(x))^{(k)}=(-1)^{k-1}{\tfrac {(k-1)!}{x^{k}}}{\text{ for all }}k\in \mathbb {N} \right.\\[0.3em]&={\binom {2016}{0}}x^{3}(-1)^{2015}{\tfrac {(2015)!}{x^{2016}}}+{\binom {2016}{1}}\cdot 3x^{2}(-1)^{2014}{\tfrac {(2014)!}{x^{2015}}}+{\binom {2016}{2}}\cdot 6x(-1)^{2013}{\tfrac {(2013)!}{x^{2014}}}+{\binom {2016}{3}}\cdot 6\cdot (-1)^{2012}{\tfrac {(2012)!}{x^{2013}}}\\[0.3em]&=-x^{3}{\tfrac {(2015)!}{x^{2016}}}+2016\cdot 3x^{2}{\tfrac {(2014)!}{x^{2015}}}-2016\cdot 2015\cdot 3x{\tfrac {(2013)!}{x^{2014}}}+2016\cdot 2015\cdot 2014\cdot {\tfrac {(2012)!}{x^{2013}}}\\[0.3em]\end{aligned}}$