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

Derivative of the inverse function

Exercise (Derivative of the inverse function)

Consider the function

$f:\mathbb {R} \to \mathbb {R} ,\ f(x)=e^{x}+x$

Justify that the following derivatives and limits exist and calculate them:

$f'(0)$ and $(f^{-1})'(1)$
$\lim _{y\to -\infty }f^{-1}(y)$ and $\lim _{y\to -\infty }(f^{-1})'(y)$

Solution (Derivative of the inverse function)

Part 1:

Proof step: Existence and computation of $f'(0)$

$f$ is differentiable on $\mathbb {R}$ as a sum of the differentiable functions $\exp$ and ${\text{id}}$ with derivative

$f'(x)=e^{x}+1$

So $f'(0)=e^{0}+1=2$ .

Proof step: Existence and computation of $(f^{-1})'(1)$

$f$ is differentiable on $\mathbb {R}$ as a sum of the differentiable functions $\exp$ and ${\text{id}}$ with derivative

$f'(x)=e^{x}+1$

Further, $f'(x)>0$ , since $e^{x}>0$ for $x\in \mathbb {R}$ . So $f$ is strictly monotonically increasing by the monotonicity criterion and hence injective. Further, $f:\mathbb {R} \to f(\mathbb {R} )$ is bijective. And we have $f(0)=1$ , so $1\in f(\mathbb {R} )$ , and there is $f'(0)=2\neq 0$ . Hence $f^{-1}$ is differentiable at $1$ with

$(f^{-1})'(1)={\frac {1}{f'(0)}}={\frac {1}{2}}$

Part 2:

Proof step: Computation of $\lim _{y\to -\infty }f^{-1}(y)$

There is

$\lim _{x\to -\infty }f(x)=\lim _{x\to -\infty }\underbrace {e^{x}} _{\to 0}+\underbrace {x} _{\to -\infty }=-\infty$

So

$\lim _{y\to -\infty }f^{-1}(y)=-\infty$

Proof step: Computation of $\lim _{y\to -\infty }(f^{-1})'(y)$

There is $\lim _{x\to \infty }f(x)=\infty$ . According to the intermediate value theorem, $f:\mathbb {R} \to \mathbb {R}$ is bijective. For all $y\in \mathbb {R}$ we thus have

$(f^{-1})'(y)={\frac {1}{f'(f^{-1}(y))}}={\frac {1}{\exp(f^{-1}(y))+1}}$

So

$\lim _{y\to -\infty }(f^{-1})'(y)=\lim _{y\to -\infty }{\frac {1}{\exp(f^{-1}(y))+1}}={\frac {1}{\exp(\lim _{y\to -\infty }f^{-1}(y))+1}}={\frac {1}{1}}=1$

Derivative of a general logarithm function

Exercise (Derivative of a general logarithm function)

Show that the function

$f:\mathbb {R} \to \mathbb {R} ^{+},\ f(x)=a^{x}=\exp(x\ln a)$

with base $a\in \mathbb {R} \setminus \{1\}$ is bijective and differentiable on $\mathbb {R}$ . Show further that the inverse function

$f^{-1}=\log _{a}:\mathbb {R} ^{+}\to \mathbb {R} ,\ \log _{a}(y)={\frac {\ln y}{\ln a}}$

is differentiable on all of $\mathbb {R} ^{+}$ with derivative $(\log _{a})'(y)={\frac {1}{y\ln a}}$ .

Solution (Derivative of a general logarithm function)

Proof step: $f$ is bijective and differentiable

Case 1: $a>1$

$f$ is continuous on $\mathbb {R}$ as a composition of continuous functions. Further , since $\ln a>0$ , $\lim _{x\to \infty }\exp(x)=\infty$ and $\lim _{x\to -\infty }\exp(x)=0$ there is:

$\lim _{x\to \infty }a^{x}=\lim _{x\to \infty }\exp(x\ln a)=\infty$

as well as

$\lim _{x\to -\infty }a^{x}=\lim _{x\to -\infty }\exp(x\ln a)=0$

By the intermediate value theorem, $f(\mathbb {R} )=\mathbb {R} ^{+}$ and $f$ is therefore surjective. Moreover, $f$ is differentiable according to the chain rule as a composition of differentiable functions, and

$f'(x)=\underbrace {\ln a} _{>0}\cdot \underbrace {\exp(x\ln a)} _{>0}>0$

for all $x\in \mathbb {R}$ . By the monotonicity criterion, $f$ is strictly monotonically increasing, and hence injective. So we have shown the bijectivity of $f$ .

Case 2: $0<a<1$

$f$ is continuous on $\mathbb {R}$ as a composition of continuous functions. Further , since $\ln a<0$ , $\lim _{x\to \infty }\exp(x)=\infty$ and $\lim _{x\to -\infty }\exp(x)=0$ there is:

$\lim _{x\to \infty }a^{x}=\lim _{x\to \infty }\exp(x\ln a)=0$

as well as

$\lim _{x\to -\infty }a^{x}=\lim _{x\to -\infty }\exp(x\ln a)=\infty$

By the intermediate value theorem, $f(\mathbb {R} )=\mathbb {R} ^{+}$ and $f$ is therefore surjective. Moreover, $f$ is differentiable according to the chain rule as a composition of differentiable functions, and

$f'(x)=\underbrace {\ln a} _{<0}\cdot \underbrace {\exp(x\ln a)} _{>0}<0$

for all $x\in \mathbb {R}$ . By the monotonicity criterion, $f$ is strictly monotonically decreasing, and hence injective. So we have shown the bijectivity of $f$ .

Proof step: $f^{-1}$ exists and is differentiable

$f$ is bijective on $\mathbb {R}$ and hence invertible. Further, there is

$y=\exp(x\ln a)\iff \ln y=x\ln a\iff x={\frac {\ln y}{\ln a}}$

So

$f^{-1}:\mathbb {R} ^{+}\to \mathbb {R} ,f^{-1}(y)=\log _{a}(y)={\frac {\ln y}{\ln a}}$

Since in addition $f'(x)=\ln a\cdot \exp(x\ln a)\neq 0$ for all $x=f^{-1}(y)\in \mathbb {R}$ , the theorem about inverse function derivatives yields that $f^{-1}$ is differentiable on $\mathbb {R} ^{+}$ with

$(f^{-1})'(y)={\frac {1}{f'(f^{-1}(y))}}={\frac {1}{f'({\tfrac {\ln y}{\ln a}})}}={\frac {1}{\ln a\cdot \exp({\tfrac {\ln y}{\ln a}}\ln a)}}={\frac {1}{\ln a\cdot \exp(\ln y)}}={\frac {1}{\ln a\cdot y}}$

Exercise (derivatives on

\operatorname {arccot}

and area-functions)

Show that the functions

$\operatorname {arccot} :\mathbb {R} \to (0,\pi )$
${\text{arsinh}}:\mathbb {R} \to \mathbb {R}$
${\text{arcosh}}:(1,\infty )\to (0,\infty )$
${\text{artanh}}:(-1,1)\to \mathbb {R}$

are differentiable, and determine their derivative.

Proof (derivatives on

\operatorname {arccot}

and area-functions)

Differentiability of $\operatorname {arccot}$ :

For the cotangent function $\cot |_{(0,\pi )}=\left.{\tfrac {\cos }{\sin }}\right|_{(0,\pi )}$ there is: $\cot '=-1-\cot ^{2}<0$ . Thus the function is differentiable and strictly monotonically decreasing, and thus injective. Further, $\cot((0,\pi ))=\mathbb {R}$ . Thus $\cot :(0,\pi )\to \mathbb {R}$ is bijective. The inverse function

$\operatorname {arccot} :\mathbb {R} \to (0,\pi )$

is differentiable according to the theorem on the derivative of the inverse function, and for ${\tilde {x}}\in \mathbb {R}$ there is:

$\operatorname {arccot} '({\tilde {x}})={\frac {1}{\cot '(\operatorname {arccot}({\tilde {x}}))}}={\frac {1}{-1-\cot ^{2}(\operatorname {arccot}({\tilde {x}}))}}={\frac {1}{-1-{\tilde {x}}^{2}}}=-{\frac {1}{1+{\tilde {x}}^{2}}}$

Differentiability of ${\text{arsinh}}$ :

The hyperbolic sine function $\sinh :\mathbb {R} \to \mathbb {R}$ is differentiable with $\sinh '=\cosh >0$ . Thus it is strictly monotonically increasing, and hence injective. Further, $\sinh(\mathbb {R} )=\mathbb {R}$ . So it is also surjective. The inverse function

${\text{arsinh}}:\mathbb {R} \to \mathbb {R}$

is differentiable by the theorem on the derivative of the inverse function , and for ${\tilde {x}}\in \mathbb {R}$ there is:

${\text{arsinh}}'({\tilde {x}})={\frac {1}{\sinh '({\text{arsinh}}({\tilde {x}}))}}={\frac {1}{\cosh({\text{arsinh}}({\tilde {x}}))}}={\frac {1}{\sqrt {\sinh ^{2}({\text{arsinh}}({\tilde {x}}))+1}}}={\frac {1}{\sqrt {{\tilde {x}}^{2}+1}}}$

Differentiability of ${\text{arcosh}}$ :

The hyperbolic cosine function $\cosh :(0,\infty )\to (1,\infty )$ is differentiable with $\cosh '=\sinh >0$ on $(0,\infty )$ . Thus it is strictly monotonically increasing, and hence injective. Further, $\cosh((0,\infty ))=(1,\infty )$ . So it is also surjective. The inverse function

${\text{arcosh}}:(1,\infty )\to (0,\infty )$

is differentiable by to the theorem on the derivative of the inverse function, and for ${\tilde {x}}\in (1,\infty )$ there is:

${\text{arcosh}}'({\tilde {x}})={\frac {1}{\cosh '({\text{arcosh}}({\tilde {x}}))}}={\frac {1}{\sinh({\text{arcosh}}({\tilde {x}}))}}={\frac {1}{\sqrt {\sinh ^{2}({\text{arsinh}}({\tilde {x}}))-1}}}={\frac {1}{\sqrt {{\tilde {x}}^{2}-1}}}$

Differentiability of ${\text{artanh}}$ :

For the cotangent function $\tan :\mathbb {R} \to (-1,1)$ there is: $\tanh '=1-\tanh ^{2}>0$ . Thus the function is strictly monotonically increasing, and thus injective. Further, $\tanh(\mathbb {R} )=(-1,1)$ . Thus $\tanh :\mathbb {R} \to (-1,1)$ is bijective. The inverse function

${\text{artanh}}:(-1,1)\to \mathbb {R}$

is differentiable by the theorem on the derivative of the inverse function, and for ${\tilde {x}}\in (-1,1)$ there is:

${\text{artanh}}'({\tilde {x}})={\frac {1}{\tanh '({\text{artanh}}({\tilde {x}}))}}={\frac {1}{1-\tanh ^{2}({\text{artanh}}({\tilde {x}}))}}={\frac {1}{1-{\tilde {x}}^{2}}}$

Exercise (Non-differentiable functions at zero)

Let $\epsilon >0$ . Show that:

Let $f,g:(-\epsilon ,\epsilon )\to \mathbb {R}$ with $f(0)=g(0)=0$ and $f(x)g(x)=x$ for all $x\in (-\epsilon ,\epsilon )$ . Then $f$ and $g$ are not simultaneously differentiable at zero.
Let $f,g:(-\epsilon ,\epsilon )\to \mathbb {R}$ , and let $f$ be differentiable at zero. Further let $f(0)=f'(0)=0$ and $g(f(x))=x$ for all $x\in (-\epsilon ,\epsilon )$ . Then $g$ is not differentiable at zero.

Solution (Non-differentiable functions at zero)

Part 1: Assuming $f$ and $g$ are differentiable at zero. Then according to the product rule, $fg$ is also differentiable at zero, and from $fg=\mathrm {id}$ we get for all $x\in (-\epsilon ,\epsilon )$ that:

$f'(0)\underbrace {g(0)} _{=0}+\underbrace {f(0)} _{=0}g'(0)=1\iff 0=1$

So $f$ and $g$ cannot both be differentiable at zero.

Part 2: Assume that $g$ is differentiable at zero. Then according to the chain rule $g\circ f$ is also differentiable at zero, and from $g\circ f=\mathrm {id}$ we get for all $x\in (-\epsilon ,\epsilon )$ that:

$g'(\underbrace {f(0)} _{=0})\cdot \underbrace {f'(0)} _{=0}=1\iff 0=1$

So $g$ cannot be differentiable at zero.

Derivatives of higher order

Exercise 1

Exercise (Arbitrarily often differentiable function)

Show that the function $f:\mathbb {R} \to \mathbb {R} ,\ f(x)=xe^{x}$ is arbitrarily often differentiable and for all $n\in \mathbb {N}$ there is:

$f^{(n)}(x)=(x+n)e^{x}$

Proof (Arbitrarily often differentiable function)

The proof goes by induction over $n\in \mathbb {N}$ :

Theorem whose validity shall be proven for $m\in \mathbb {N}$ bewiesen werden soll:

$f^{(n)}(x)=(x+n)e^{x}$

1. Base case:

$f'(x){\underset {\text{rule}}{\overset {\text{product}}{=}}}1\cdot e^{x}+x\cdot e^{x}=(x+1)e^{x}$

2. Inductive step:

2a. Inductive hypothesis:

$f^{(n)}(x)=(x+n)e^{x}$

2b. Induction theorem:

$f^{(n+1)}(x)=(x+n+1)e^{x}$

2c. Proof of induction step:

$f^{(n+1)}(x)=(f^{(n)}(x))'{\underset {\text{assumption}}{\overset {\text{induction}}{=}}}\left((x+n)e^{x}\right)'{\underset {\text{rule}}{\overset {\text{product}}{=}}}1\cdot e^{x}+(x+n)e^{x}=(x+n+1)e^{x}$

Exercise 2

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

Provide an example of a

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

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

Solution sub-exercise 1:

$f(x)=|x|$

Solution sub-exercise 2:

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

Solution sub-exercise 3:

$f(x)=x\cdot |x|$

Solution sub-exercise 4:

$f(x)=x^{2}\cdot |x|$

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

$f(x)=x^{k}\cdot |x|$

Exercise 3

Exercise (Application of the Leibniz rule)

Determine the following derivatives using the Leibniz rule

$(x\sin(x))^{(10)}$
$(x^{3}\ln(x))^{(2016)}$ 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

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