Exercises: Derivatives 1 – Serlo

Computing derivatives with differential quotients

Exrecises: derivative and differentiability

Exercise (Differentiable power function)

Show that the power function $f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-2}={\tfrac {1}{x^{2}}}$ is differentiable at $\xi =-2$ and compute the derivative. What is the derivative of $f$ at any ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$?

Solution (Differentiable power function)

The differential quotient of $f$ at $\xi =-2$ is given by

${\begin{aligned}f'(-2)&=\lim _{h\to 0}{\frac {f(-2+h)-f(-2)}{h}}=\lim _{h\to 0}{\frac {{\frac {1}{(-2+h)^{2}}}-{\frac {1}{(-2)^{2}}}}{h}}\\[0.3em]&=\lim _{h\to 0}{\frac {\frac {(-2)^{2}-(-2+h)^{2}}{(-2)^{2}\cdot (-2+h)^{2}}}{h}}=\lim _{h\to 0}{\frac {4-(4-4h+h^{2})}{4h(h^{2}-4h+4)}}\\[0.3em]&=\lim _{h\to 0}{\frac {4h-h^{2}}{4h^{3}-16h^{2}+16h}}=\lim _{h\to 0}{\frac {4-h}{4h^{2}-16h+16}}\\[0.3em]&={\frac {4-0}{0+0+16}}={\frac {4}{16}}={\frac {1}{4}}\end{aligned}}$

So $f$ is differentiable at $\xi =-2$ , with derivative $f'(-2)={\tfrac {1}{4}}$. For a general ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is

${\begin{aligned}f'({\tilde {x}})&=\lim _{h\to 0}{\frac {f({\tilde {x}}+h)-f({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {{\frac {1}{({\tilde {x}}+h)^{2}}}-{\frac {1}{{\tilde {x}}^{2}}}}{h}}\\[0.3em]&=\lim _{h\to 0}{\frac {\frac {{\tilde {x}}^{2}-({\tilde {x}}+h)^{2}}{{\tilde {x}}^{2}\cdot ({\tilde {x}}+h)^{2}}}{h}}=\lim _{h\to 0}{\frac {{\tilde {x}}^{2}-({\tilde {x}}^{2}+2{\tilde {x}}h+h^{2})}{{\tilde {x}}^{2}h({\tilde {x}}^{2}+2{\tilde {x}}h+h^{2})}}\\[0.3em]&=\lim _{h\to 0}{\frac {-2{\tilde {x}}h-h^{2}}{{\tilde {x}}^{4}h+2{\tilde {x}}^{3}h^{2}+{\tilde {x}}^{2}h^{3}}}=\lim _{h\to 0}{\frac {-2{\tilde {x}}-h}{{\tilde {x}}^{4}+2{\tilde {x}}^{3}h+{\tilde {x}}^{2}h^{2}}}\\[0.3em]&={\frac {-2{\tilde {x}}-0}{{\tilde {x}}^{4}+0+0}}=-{\frac {2}{{\tilde {x}}^{3}}}\end{aligned}}$

Exercise (Derivative of a product function)

Let $f:\mathbb {R} \to \mathbb {R}$ defined by

$f(x)=x(x-1)(x-2)\cdot \ldots \cdot (x-1000)=\prod _{k=0}^{1000}(x-k)$

Determine $f'(0)$.

Solution (Derivative of a product function)

There is

$f'(0)=\lim _{x\to 0}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0}{\frac {\prod _{k=0}^{1000}(x-k)-0}{x-0}}=\lim _{x\to 0}{\frac {x\cdot \prod _{k=1}^{1000}(x-k)}{x}}=\lim _{x\to 0}\prod _{k=1}^{1000}(x-k){\overset {(*)}{=}}\prod _{k=1}^{1000}(-k)=\underbrace {(-1)^{1000}} _{=1}\cdot \underbrace {\prod _{k=1}^{1000}k} _{=1000!}=1000!$

For $(*)$ we have used that $x\mapsto \prod _{k=1}^{1000}(x-k)$ is continuous as a product of the continuous functions $x\mapsto x-k$ for $1\leq k\leq 1000$.

Exercise (Derivative of a function with case distinction)

Check whether the following functions are differentiable at $x=0$.

1. $f:\mathbb {R} \to \mathbb {R} ,\ f(x)={\begin{cases}\cos({\tfrac {1}{x}})&{\text{ for }}x\neq 0\\0&{\text{ for }}x=0\end{cases}}$
2. $g:\mathbb {R} \to \mathbb {R} ,\ g(x)={\begin{cases}x\cos({\tfrac {1}{x}})&{\text{ for }}x\neq 0\\0&{\text{ for }}x=0\end{cases}}$

Solution (Derivative of a function with case distinction)

Part 1: Since $\cos({\tfrac {1}{x}})$ oscillates very quickly between $-1$ and $-1$, just like $\sin({\tfrac {1}{x}})$ for $x\to 0$, it is to be expected that $f$ at $x=0$ is not continuous. For this purpose we consider the null sequences $(a_{n})_{n\in \mathbb {N} }=({\tfrac {1}{2n\pi }})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }=({\tfrac {1}{(2n-1)\pi }})_{n\in \mathbb {N} }$. For these sequences

$f(a_{n})=\cos(2n\pi )=1$

and

$f(b_{n})=\cos((2n-1)\pi )=-1$

So $\lim _{x\to 0}f(x)$ does not exist. According to the sequence criterion $f$ is therefore not continuous at zero and thus not differentiable.

Part 2: The function $g$ is continuous at zero by the sequence criterion, as $\lim _{x\to 0}g(x)=0$. So we can consider the differential quotient:

$\lim _{x\to 0}{\frac {g(x)-g(0)}{x-0}}=\lim _{x\to 0}{\frac {x\cos({\tfrac {1}{x}})-0}{x}}=\lim _{x\to 0}\cos({\tfrac {1}{x}})$

In Part 1 we have shown that this limit value does not exist. Therefore $g$ is also not differentiable at zero.

Exercise (Criterion for non-differentiability of a general function at zero)

Let $f:(-1,1)\to \mathbb {R}$. Show that: If $|f(x)|\geq |x|^{\alpha }$ for some $0<\alpha <1$ and $f(0)=0$, then $f$ is not differentiable at zero.

Solution (Criterion for non-differentiability of a general function at zero)

There is

$\left|{\frac {f(x)-f(0)}{x-0}}\right|={\frac {|f(x)|}{|x|}}{\overset {|f(x)|\geq |x|^{\alpha }}{\geq }}{\frac {|x|^{\alpha }}{|x|}}={\frac {1}{|x|^{1-\alpha }}}{\overset {x\to 0}{\longrightarrow }}\infty$ since $1-\alpha >0$

So $f'(0)=\lim _{x\to 0}{\frac {f(x)-f(0)}{x-0}}$ does not exist.

Exercise (Determining limits with the differential quotient)

Let $f:D\to \mathbb {R}$ be differentiable at $a\in D$. Show that the following limits exist

1. $\lim _{h\to 0}{\frac {f(a+h^{2})-f(a)}{h}}=0$
2. $\lim _{h\to 0}{\frac {f(a+ch)-f(a-dh)}{h}}=(c+d)f'(a)$ for $c,d\in \mathbb {R}$
3. $\lim _{x\to a}{\frac {xf(a)-af(x)}{x-a}}=f(a)-af'(a)$

How to get to the proof? (Determining limits with the differential quotient)

Since $f$ in $a$ is differentiable, there is

$\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}=f'(a)$ and $\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}=f'(a)$

In addition, from the exercises within the article "derivatives", we know that

$\lim _{h\to 0}{\frac {f(a-h)-f(a)}{h}}=-f'(x)$

The idea is to transform the limits so that we can calculate them using the differential quotient.

Solution (Determining limits with the differential quotient)

Part 1: Because $\lim _{h\to 0}{\tfrac {f(a+h)-f(a)}{h}}=f'(a)$ there is also $\lim _{h\to 0}{\tfrac {f(a+h^{2})-f(a)}{h^{2}}}=f'(a)$. So

$\lim _{h\to 0}{\frac {f(a+h^{2})-f(a)}{h}}=\lim _{h\to 0}h\cdot {\frac {f(a+h^{2})-f(a)}{h^{2}}}=\underbrace {\lim _{h\to 0}h} _{=0}\cdot \underbrace {\lim _{h\to 0}{\frac {f(a+h^{2})-f(a)}{h^{2}}}} _{=f'(a)}=0$

Part 2: With $\lim _{h\to 0}{\tfrac {f(a+h)-f(a)}{h}}=f'(a)$ and $\lim _{h\to 0}{\tfrac {f(a-h)-f(a)}{h}}=-f'(a)$ there is also $\lim _{h\to 0}{\tfrac {f(a+ch)-f(a)}{ch}}=f'(a)$ and $\lim _{h\to 0}{\tfrac {f(a-dh)-f(a)}{dh}}=-f'(a)$. Hence

${\begin{aligned}\lim _{h\to 0}{\frac {f(a+ch)-f(a-dh)}{h}}&=\lim _{h\to 0}{\frac {f(a+ch)-f(a)-(f(a-dh)-f(a))}{h}}=\lim _{h\to 0}{\frac {f(a+ch)-f(a)}{h}}-\lim _{h\to 0}{\frac {f(a-dh)-f(a)}{h}}\\&=c\cdot \underbrace {\lim _{h\to 0}{\frac {f(a+ch)-f(a)}{ch}}} _{=f'(a)}-d\cdot \underbrace {\lim _{h\to 0}{\frac {f(a-dh)-f(a)}{dh}}} _{=-f'(a)}=cf'(a)+df'(a)=(c+d)f'(a)\end{aligned}}$

Part 3: Here we need the "original" differential quotient $\lim _{x\to a}{\tfrac {f(x)-f(a)}{x-a}}=f'(a)$:

${\begin{aligned}\lim _{x\to a}{\frac {xf(a)-af(x)}{x-a}}&=\lim _{x\to a}{\frac {xf(a)-af(a)-(af(x)-af(a))}{x-a}}=\lim _{x\to a}{\frac {xf(a)-af(a)}{x-a}}-\lim _{x\to a}{\frac {af(x)-af(a)}{x-a}}\\&=\lim _{x\to a}{\frac {(x-a)f(a)}{x-a}}-\lim _{x\to a}{\frac {a(f(x)-f(a))}{x-a}}=c\cdot \underbrace {\lim _{x\to a}f(a)} _{=f(a)}-a\cdot \underbrace {\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}} _{=f'(a)}=f(a)-af'(a)\end{aligned}}$

Exercise (Implication of differentiability)

Let $f:\mathbb {R} \to \mathbb {R}$ be differentiable at ${\tilde {x}}$. Further let $(a_{n})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }$ be sequences with $a_{n}<{\tilde {x}}<b_{n}$ for all $n\in \mathbb {N}$, as well as $\lim _{n\to \infty }(a_{n}-b_{n})=0$. Show that then, there is

$\lim _{n\to \infty }{\frac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}=f'({\tilde {x}})$

Additional question: Does the converse statement also hold? I.e. does the limit value $\lim _{n\to \infty }{\tfrac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}$ with sequences $(a_{n})$ and $(b_{n})$ as above exist, so $f$ is differentiable at ${\tilde {x}}$, and $f'({\tilde {x}})$ is equal to this limit?

Hint: Show first that ${\tfrac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}={\tfrac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}+{\tfrac {a_{n}-{\tilde {x}}}{a_{n}-b_{n}}}\left({\tfrac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}-{\tfrac {f(b_{n})-f({\tilde {x}})}{b_{n}-{\tilde {x}}}}\right)$

Solution (Implication of differentiability)

There is

${\begin{aligned}{\frac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}&={\frac {f(a_{n})-f({\tilde {x}})+f({\tilde {x}})-f(b_{n})}{a_{n}-b_{n}}}\\[0.3em]&={\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-b_{n}}}+{\frac {f({\tilde {x}})-f(b_{n})}{a_{n}-b_{n}}}\\[0.3em]&={\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}+{\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-b_{n}}}-{\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}-{\frac {f(b_{n})-f({\tilde {x}})}{a_{n}-b_{n}}}\\[0.3em]&={\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}+{\frac {(a_{n}-{\tilde {x}}-(a_{n}-b_{n}))(f(a_{n})-f({\tilde {x}})}{(a_{n}-b_{n})(a_{n}-{\tilde {x}})}}-{\frac {f(b_{n})-f({\tilde {x}})}{a_{n}-b_{n}}}\\[0.3em]&={\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}+{\frac {b_{n}-{\tilde {x}}}{a_{n}-b_{n}}}\cdot {\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}-{\frac {b_{n}-{\tilde {x}}}{a_{n}-b_{n}}}\cdot {\frac {f(b_{n})-f({\tilde {x}})}{b_{n}-{\tilde {x}}}}\\[0.3em]&=\underbrace {\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}} _{\to f'({\tilde {x}})}+\underbrace {\frac {a_{n}-{\tilde {x}}}{a_{n}-b_{n}}} _{|\ldots |\leq 1}\cdot \underbrace {\left({\frac {f(a_{n})-f({\tilde {x}})}{a_{n}-{\tilde {x}}}}-{\frac {f(b_{n})-f({\tilde {x}})}{b_{n}-{\tilde {x}}}}\right)} _{\to f'({\tilde {x}})-f'({\tilde {x}})=0}\end{aligned}}$

Since now the product of a bounded sequence and a null sequence converges to zero, there is with the calculation rules for sequences

$\lim _{n\to \infty }{\frac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}=f'({\tilde {x}})+0=f'({\tilde {x}})$

Concerning the additional question: The converse is false. Let us consider the following function, which is not continuous at ${\tilde {x}}=0$ (and therefore not differentiable):

$f:\mathbb {R} \to \mathbb {R} ,\ f(x)={\begin{cases}1&{\text{ for }}x\neq 0,\\0&{\text{ for }}x=0\end{cases}}$

Then, there is for all null sequences $(a_{n})$ and $(b_{n})$ with $a_{n}<0<b_{n}$:

$\lim _{n\to \infty }{\frac {f(a_{n})-f(b_{n})}{a_{n}-b_{n}}}=\lim _{n\to \infty }{\frac {1-1}{a_{n}-b_{n}}}=\lim _{n\to \infty }{\frac {0}{a_{n}-b_{n}}}=0$

Exercises: examples for derivatives

Exercise (Derivatives of linear and quadratic functions)

Determine (using the definition) the derivative of a linear function

$f:\mathbb {R} \to \mathbb {R} ,\ f(x)=ax+b$

and of a quadratic function

$g:\mathbb {R} \to \mathbb {R} ,\ g(x)=ax^{2}+bx+c$

with $a,b,c\in \mathbb {R}$.

Solution (Derivatives of linear and quadratic functions)

1. linear function: For ${\tilde {x}}\in \mathbb {R}$ there is

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {ax+b-(a{\tilde {x}}+b)}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {ax+b-a{\tilde {x}}-b}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {ax-a{\tilde {x}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {a(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}a=a$

2. quadratic function: For ${\tilde {x}}\in \mathbb {R}$ there is

${\begin{aligned}g'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {g(x)-g({\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {ax^{2}+bx+c-(a{\tilde {x}}^{2}+b{\tilde {x}}+c)}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {ax^{2}+bx-a{\tilde {x}}^{2}-b{\tilde {x}}}{x-{\tilde {x}}}}\\&=\lim _{x\to {\tilde {x}}}{\frac {a(x^{2}-{\tilde {x}}^{2})+b(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {a(x+{\tilde {x}})(x-{\tilde {x}})}{x-{\tilde {x}}}}+\lim _{x\to {\tilde {x}}}{\frac {b(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}a(x+{\tilde {x}})+\lim _{x\to {\tilde {x}}}b=2a{\tilde {x}}+b\end{aligned}}$

Exercise (Derivative of the logarithm function)

Compute the derivative of the natural logarithm function

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

directly, using the differential quotient.

Solution (Derivative of the logarithm function)

1st way:

For ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\ln x-\ln {\tilde {x}}}{x-{\tilde {x}}}}$

Nun for $0<x<{\tilde {x}}$ we have the inequality

${\frac {1}{\tilde {x}}}\leq {\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}\leq {\frac {1}{x}}$

If we swap the roles of $x$ and ${\tilde {x}}$, then there is

${\frac {1}{x}}\leq {\frac {f({\tilde {x}})-f(x)}{{\tilde {x}}-x}}\leq {\frac {1}{\tilde {x}}}$

Since the left and right-hand sides of the inequality for $x\to {\tilde {x}}$ converge towards ${\frac {1}{\tilde {x}}}$, the squeeze theorem implies

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {\ln x-\ln {\tilde {x}}}{x-{\tilde {x}}}}={\frac {1}{\tilde {x}}}$

2nd way: $h$-method

For ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

${\begin{aligned}f'({\tilde {x}})&=\lim _{h\to {\tilde {0}}}{\frac {f({\tilde {x}}+h)-f({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\ln({\tilde {x}}+h)-\ln({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\ln({\tfrac {{\tilde {x}}+h}{\tilde {x}}})}{h}}=\lim _{h\to 0}{\frac {\ln(1+{\tfrac {h}{\tilde {x}}})}{h}}=\lim _{h\to 0}{\frac {1}{\tilde {x}}}{\frac {\ln(1+{\tfrac {h}{\tilde {x}}})}{\frac {h}{\tilde {x}}}}\\&{\overset {y={\frac {h}{\tilde {x}}}}{=}}\lim _{y\to 0}{\frac {1}{\tilde {x}}}{\frac {\ln(1+y)}{y}}={\frac {1}{\tilde {x}}}\underbrace {\lim _{y\to 0}{\frac {\ln(1+y)}{y}}} _{=1}={\frac {1}{\tilde {x}}}\end{aligned}}$

Exercise (Computing the derivatives of hyperbolic functions $\sinh$ $\cosh$ and $\tanh$)

Determine die derivatives of the following functions using the differential quotient

1. $\sinh :\mathbb {R} \to \mathbb {R} ,\sinh(x):={\frac {e^{x}-e^{-x}}{2}}$
2. $\cosh :\mathbb {R} \to \mathbb {R} ,\cosh(x):={\frac {e^{x}+e^{-x}}{2}}$
3. $\tanh :\mathbb {R} \to \mathbb {R} ,\tanh(x):={\frac {\sinh }{\cosh }}$

Solution (Computing the derivatives of hyperbolic functions $\sinh$ $\cosh$ and $\tanh$)

Part 1: Let $x\in \mathbb {R}$. Then, there is

${\begin{aligned}&\sinh '(x)=\lim _{h\to 0}{\frac {\sinh(x+h)-\sinh(x)}{h}}=\ &\lim _{h\to 0}{\frac {{\tfrac {e^{x+h}-e^{-(x+h)}}{2}}-{\frac {e^{x}-e^{-x}}{2}}}{h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)-e^{-x}(e^{-h}-1)}{2h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)}{2h}}-\lim _{h\to 0}{\frac {e^{-x}(e^{-h}-1)}{2h}}\\[0.3em]&\color {Gray}\left\downarrow \ {\text{substitution (2nd limit) }}{\tilde {h}}=-h\right.\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)}{2h}}-\lim _{{\tilde {h}}\to 0}{\frac {e^{-x}(e^{\tilde {h}}-1)}{-2{\tilde {h}}}}\\[0.3em]=\ &{\frac {e^{x}}{2}}\lim _{h\to 0}{\frac {(e^{h}-1)}{h}}+{\frac {e^{-x}}{2}}\lim _{{\tilde {h}}\to 0}{\frac {(e^{\tilde {h}}-1)}{\tilde {h}}}\\[0.3em]&\color {Gray}\left\downarrow \ \lim _{h\to 0}{\frac {(e^{h}-1)}{h}}=1\right.\\[0.3em]=\ &{\frac {e^{x}+e^{-x}}{2}}\\[0.3em]=\ &\cosh(x)\end{aligned}}$

Alternative proof:

${\begin{aligned}\sinh '(x)&=\lim _{h\to 0}{\frac {\sinh(x+h)-\sinh(x)}{h}}\\[0.3em]&\color {Gray}\left\downarrow \ {\text{addition theorem: }}\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)\right.\\[0.3em]&=\lim _{h\to 0}{\frac {\sinh(x)\cosh(h)+\cosh(x)\sinh(h)-\sinh(x)}{h}}\\[0.3em]&=\sinh(x)\cdot \underbrace {\lim _{h\to 0}{\frac {\cosh(h)-1}{h}}} _{=0}+\cosh(x)\cdot \underbrace {\lim _{h\to 0}{\frac {\sinh(h)}{h}}} _{=1}\\[0.3em]&=\cosh(x)\end{aligned}}$

Part 2: Let $x\in \mathbb {R}$. Then, there is

${\begin{aligned}&\cosh '(x)=\lim _{h\to 0}{\frac {\cosh(x+h)-\cosh(x)}{h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {{\tfrac {e^{x+h}+e^{-(x+h)}}{2}}-{\tfrac {e^{x}+e^{-x}}{2}}}{h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)+e^{-x}(e^{-h}-1)}{2h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)}{2h}}+\lim _{h\to 0}{\frac {e^{-x}(e^{-h}-1)}{2h}}\\[0.3em]&\color {Gray}\left\downarrow \ {\text{substitution (2nd limit) }}{\tilde {h}}=-h\right.\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x}(e^{h}-1)}{2h}}+\lim _{{\tilde {h}}\to 0}{\frac {e^{-x}(e^{\tilde {h}}-1)}{-2{\tilde {h}}}}\\[0.3em]=\ &{\frac {e^{x}}{2}}\lim _{h\to 0}{\frac {(e^{h}-1)}{h}}-{\frac {e^{-x}}{2}}\lim _{{\tilde {h}}\to 0}{\frac {(e^{\tilde {h}}-1)}{\tilde {h}}}\\[0.3em]&\color {Gray}\left\downarrow \ \lim _{h\to 0}{\frac {(e^{h}-1)}{h}}=1\right.\\[0.3em]=\ &{\frac {e^{x}-e^{-x}}{2}}\\[0.3em]=\ &\sinh(x)\end{aligned}}$

Alternative proof:

${\begin{aligned}\cosh '(x)&=\lim _{h\to 0}{\frac {\cosh(x+h)-\cosh(x)}{h}}\\[0.3em]&\color {Gray}\left\downarrow \ {\text{addition theorem: }}\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y)\right.\\[0.3em]&=\lim _{h\to 0}{\frac {\cosh(x)\cosh(h)+\sinh(x)\sinh(h)-\cosh(x)}{h}}\\[0.3em]&=\cosh(x)\cdot \underbrace {\lim _{h\to 0}{\frac {\cosh(h)-1}{h}}} _{=0}+\sinh(x)\cdot \underbrace {\lim _{h\to 0}{\frac {\sinh(h)}{h}}} _{=1}\\[0.3em]&=\sinh(x)\end{aligned}}$

Part 3: Let $x\in \mathbb {R}$. Then, there is

$\tanh(x)={\frac {\sinh(x)}{\cosh(x)}}={\frac {\frac {e^{x}-e^{-x}}{2}}{\frac {e^{x}+e^{-x}}{2}}}={\frac {2(e^{x}-e^{-x})}{2(e^{x}+e^{-x})}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}$

So

${\begin{aligned}&\tanh '(x)=\lim _{h\to 0}{\frac {\tanh(x+h)-\tanh(x)}{h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {{\frac {e^{x+h}-e^{-(x+h)}}{e^{x+h}+e^{-(x+h)}}}-{\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}}{h}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {(e^{x+h}-e^{-(x+h)})(e^{x}+e^{-x})-(e^{x}-e^{-x})(e^{x+h}+e^{-(x+h)})}{h(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {e^{x+h}e^{x}-e^{-x-h}e^{x}+e^{-x}e^{x+h}-e^{-x}e^{-x-h}-e^{x}e^{x+h}-e^{x}e^{-x-h}+e^{-x}e^{x+h}+e^{-x}e^{-x-h}}{h(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {2e^{-x}e^{x+h}-2e^{x}e^{-x-h}}{h(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {2e^{h}-2e^{-h}}{h(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]=\ &\lim _{h\to 0}{\frac {2e^{-h}(e^{2h}-1)}{h(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]=\ &\lim _{h\to 0}{2e^{-h}\cdot {\frac {(e^{2h}-1)}{h}}\cdot {\frac {1}{(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}}\\[0.3em]=\ &\lim _{h\to 0}{2e^{-h}\cdot 2{\frac {(e^{2h}-1)}{2h}}\cdot {\frac {1}{(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}}\\[0.3em]=\ &\lim _{h\to 0}2e^{-h}\cdot \lim _{h\to 0}2{\frac {(e^{2h}-1)}{2h}}\cdot \lim _{h\to 0}2e^{-h}{\frac {1}{(e^{x+h}+e^{-(x+h)})(e^{x}+e^{-x})}}\\[0.3em]&\color {Gray}\left\downarrow \ \lim _{h\to 0}{\frac {(e^{2h}-1)}{2h}}=1\right.\\[0.3em]=\ &2e^{0}\cdot 2\cdot {\frac {1}{(e^{x}+e^{-x})(e^{x}+e^{-x})}}\\[0.3em]=\ &{\frac {4}{(e^{x}+e^{-x})^{2}}}\\[0.3em]=\ &{\frac {1}{\left({\frac {e^{x}+e^{-x}}{2}}\right)^{2}}}\\[0.3em]=\ &{\frac {1}{\cosh ^{2}(x)}}\end{aligned}}$

Alternative proof:

${\begin{aligned}\tanh '(x)&=\lim _{h\to 0}{\frac {\tanh(x+h)-\tanh(x)}{h}}\\[0.3em]&=\lim _{h\to 0}{\frac {{\frac {\sinh(x+h)}{\cosh(x+h)}}-{\frac {\sinh(x)}{\cosh(x)}}}{h}}\\[0.3em]&=\lim _{h\to 0}{\frac {\frac {\sinh(x+h)\cosh(x)-\sinh(x)\cosh(x+h)}{\cosh(x+h)\cosh(x)}}{h}}\\[0.3em]&=\lim _{h\to 0}{\frac {\sinh(x+h)\cosh(x)-\sinh(x)\cosh(x+h)}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&\color {Gray}\left\downarrow \ {\text{addition theorems: }}\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y){\text{ and }}\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y)\right.\\[0.3em]&=\lim _{h\to 0}{\frac {(\sinh(x)\cosh(h)+\cosh(x)\sinh(h))\cosh(x)-\sinh(x)(\cosh(x)\cosh(h)+\sinh(x)\sinh(h))}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&=\lim _{h\to 0}{\frac {\sinh(x)\cosh(x)\cosh(h)+\cosh ^{2}(x)\sinh(h)-\sinh(x)\cosh(x)\cosh(h)-\sinh ^{2}(x)\sinh(h)}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&=\lim _{h\to 0}{\frac {\cosh ^{2}(x)\sinh(h)-\sinh ^{2}(x)\sinh(h)}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&=\lim _{h\to 0}{\frac {(\cosh ^{2}(x)-\sinh ^{2}(x))\sinh(h)}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&\color {Gray}\left\downarrow \ \cosh ^{2}(x)-\sinh ^{2}(x)=1\right.\\[0.3em]&=\lim _{h\to 0}{\frac {\sinh(h)}{h\cosh(x+h)\cosh(x)}}\\[0.3em]&=\lim _{h\to 0}{\frac {\sinh(h)}{h}}\cdot {\frac {1}{\cosh(x+h)\cosh(x)}}\\[0.3em]&\color {Gray}\left\downarrow \ \lim _{h\to 0}{\frac {\sinh(h)}{h}}=1\right.\\[0.3em]&={\frac {1}{\cosh ^{2}(x)}}\end{aligned}}$

Computation rules for derivatives

Applying the computation rules

Exercise (derivatives of a power function)

Show by induction in $n\in \mathbb {N}$, that the power function

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

is differentiable with

$f'(x)=nx^{n-1}$

Proof (derivatives of a power function)

Induction base: $n=1$

If $f(x)=x^{1}=x$, then there is

$f'(x)=1=1\cdot x^{0}$

Induction assumption:

For $f(x)=x^{n}$ with $n\in \mathbb {N}$, there is

$f'(x)=n\cdot x^{n-1}$

Induction step: $n\to n+1$

Let $f(x)=x^{n+1}=x\cdot x^{n}$. Then $f$ is differentiable by induction assumption and the product rule. For $x\in \mathbb {R}$ there is

$f'(x){\overset {\text{IV}}{=}}1\cdot x^{n}+x\cdot nx^{n-1}=x^{n}+nx^{n}=(n+1)x^{n}$

Exercise (Derivatives of secant and cosecant)

The functions $\sec$ (secant) and $\csc$ (cosecant) are defined as follows:

$\sec(x)=(\cos x)^{-1}={\frac {1}{\cos x}}$

as well as

$\csc(x)=(\sin x)^{-1}={\frac {1}{\sin x}}$

Determine their domain of definition and all derivatives.

Solution (Derivatives of secant and cosecant)

Part 1: secant

domain of definition: $\sec$ is well-defined $\iff$ $\cos x\neq 0$ $\iff$ $x\neq {\tfrac {\pi }{2}}+k\pi$ $\forall k\in \mathbb {Z}$ $\Longrightarrow$ $D_{1}=\mathbb {R} \setminus \{{\tfrac {\pi }{2}}+k\pi \mid k\in \mathbb {Z} \}$

Derivative: Since $\cos$ is differentiable on $D_{1}$ , there is with the chain rule

$\sec(x)=(-1)(\cos x)^{-2}\cdot (-\sin x)={\frac {\sin x}{\cos ^{2}x}}=\sec x\cdot \tan x={\frac {\sec ^{2}x}{\csc x}}$

Part 2: cosecant

domain of definition: $\csc$ is well-defined $\iff$ $\sin x\neq 0$ $\iff$ $x\neq k\pi$ $\forall k\in \mathbb {Z}$ $\Longrightarrow$ $D_{2}=\mathbb {R} \setminus \{k\pi \mid k\in \mathbb {Z} \}$

Derivative: Since $\sin$ is differentiable on $D_{2}$ , there is with the chain rule

$\csc(x)=(-1)(\sin x)^{-2}\cdot (\cos x)=-{\frac {\cos x}{\sin ^{2}x}}=\csc x\cdot \cot x=-{\frac {\csc ^{2}x}{\sec x}}$

Exercise (Computing derivatives)

Determine the domain of definition of the following functions, as well as their derivatives

1. $f_{1}(x)=x^{2}e^{-x}$
2. $f_{2}(x)=\ln(\ln(\ln(x)))$
3. $f_{3}(x)=\sin(x\cos(x^{2}))$
4. $f_{4}(x)={\frac {1+x^{2}}{\sqrt {x^{2}-1}}}$
5. $f_{5}(x)=x\cdot |x|$

Solution (Computing derivatives)

Part 1:

domain of definition: $\mathbb {R}$

Derivative: For $x\in \mathbb {R}$ there is with the product rule

$f_{1}'(x)=2xe^{-x}+x^{2}(-e^{-x})=xe^{-x}(2-x)$

Part 2:

domain of definition: $(e,\infty )$, as

$\ln(\ln(x))>0\iff \ln(x)>1\iff x>e$

Derivative: For $x\in (0,\infty )$ there is with the chain rule

$f_{2}'(x)={\frac {1}{\ln(\ln(x))}}\cdot {\frac {1}{\ln(x)}}\cdot {\frac {1}{x}}={\frac {1}{x\ln(x)\ln(\ln(x))}}$

Part 3:

domain of definition: $\mathbb {R}$

Derivative: For $x\in \mathbb {R}$ there is with the chain- and product rule

$f_{3}'(x)=\cos(x\cos(x^{2}))[1\cdot \cos(x^{2})+x(-\sin(x^{2})\cdot 2x)]=\cos(x\cos(x^{2}))[\cos(x^{2})-2x^{2}\sin(x^{2})]$

Part 4:

domain of definition: $(-\infty ,-1)\cup (1,\infty )$, since there must be

1. ${\sqrt {x^{2}-1}}\neq 0\iff x^{2}\neq 1\iff x\neq \pm 1$
2. $x^{2}-1>0\iff x^{2}>1\iff |x|>1\iff (x<-1)\vee (x>1)$

Derivative: For $x\in \mathbb {R}$ there is with the quotient rule

$f_{4}'(x)={\frac {2x{\sqrt {x^{2}-1}}-(1+x^{2}){\frac {2x}{2{\sqrt {x^{2}-1}}}}}{x^{2}-1}}={\frac {2x(x^{2}-1)-x(1+x^{2})}{(x^{2}-1)^{\frac {3}{2}}}}={\frac {x(2x^{2}-2-1-x^{2})}{(x^{2}-1)^{\frac {3}{2}}}}={\frac {x(x^{2}-3)}{(x^{2}-1)^{\frac {3}{2}}}}$

Part 5:

domain of definition: $\mathbb {R}$

Derivative:

For $x<0$ there is $f_{5}(x)=x^{2}$ $\Longrightarrow$ $f_{5}'(x)=2x$

For $x<0$ there is $f_{5}(x)=-x^{2}$ $\Longrightarrow$ $f_{5}'(x)=-2x$

Further there is

$\lim _{x\to 0+}{\frac {f_{5}(x)-f_{5}(0)}{x-0}}=\lim _{x\to 0+}{\frac {x^{2}-0}{x}}=\lim _{x\to 0+}x=0$

as well as

$\lim _{x\to 0-}{\frac {f_{5}(x)-f_{5}(0)}{x-0}}=\lim _{x\to 0-}{\frac {-x^{2}-0}{x}}=\lim _{x\to 0-}-x=0$

So we have

$f_{5}'(0)=\lim _{x\to 0}{\frac {f_{5}(x)-f_{5}(0)}{x-0}}=0$

Concluding all three cases, we get for $x\in \mathbb {R}$

$f_{5}'(x)=2|x|$

Exercise (derivatives of exponential functions)

Determine the derivatives of the following functions on their domain of definition ($a>0$)

1. $f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)=x^{(x^{x})}$
2. $g:\mathbb {R} ^{+}\to \mathbb {R} ,\ g(x)=(x^{x})^{x}$
3. $h:\mathbb {R} ^{+}\to \mathbb {R} ,\ h(x)=x^{(x^{a})}$
4. $i:\mathbb {R} ^{+}\to \mathbb {R} ,\ i(x)=x^{(a^{x})}$
5. $j:\mathbb {R} ^{+}\to \mathbb {R} ,\ j(x)=a^{(x^{x})}$

For the function $g$ we may leave out the bracket, since in general $x^{x^{x}}=x^{(x^{x})}$ dis well-defined.

Solution (derivatives of exponential functions)

Part 1: There is $g(x)=x^{(x^{x})}=\exp(x^{x}\ln x)$. The function $x\mapsto x^{x}=\exp(x\ln(x))$ is differentiable with $(x^{x})'=x^{x}(\ln(x)+1)$. Hence $g$ is differentiable by the chain- and product rule and for $x\in \mathbb {R} ^{+}$ there is

$f'(x)=\exp(x^{x}\ln x)[x^{x}(\ln x+1)\cdot \ln x+x^{x}\cdot {\tfrac {1}{x}}]=\exp(x\ln x)[x^{x}(\ln x+1)\ln(x)+{\tfrac {1}{x}})]=x^{x^{x}+x}(\ln ^{2}x+\ln x+{\tfrac {1}{x}})$

Part 2: There is $g(x)=(x^{x})^{x}=x^{x\cdot x}=x^{x^{2}}=\exp(x^{2}\ln x)$. Hence $h$ is differentiable by the chain- and product rule and for $x\in \mathbb {R} ^{+}$ there is

$g'(x)=\exp(x^{2}\ln x)[2x\ln x+x^{2}\cdot {\tfrac {1}{x}}]=\exp(x^{2}\ln x)[2x\ln x+x]=\exp(x^{2}\ln x)[x(2\ln x+1)]=x^{x^{2}+1}(2\ln x+1)$

Part 3: There is $h(x)=x^{(x^{a})}=\exp(x^{a}\ln x)$. Hence $h$ is differentiable by the chain- and product rule and for $x\in \mathbb {R} ^{+}$ there is

$h'(x)=\exp(x^{a}\ln x)[ax^{a-1}\cdot \ln x+x^{a}\cdot {\tfrac {1}{x}}]=\exp(x\ln x)[x^{a-1}(a\ln x+1)]=x^{x^{a}+a-1}(a\ln x+1)$

Part 4: There is $i(x)=(x^{(a^{x})})=\exp(a^{x}\ln x)$. Hence $i$ is differentiable by the chain- and product rule and for $x\in \mathbb {R} ^{+}$ there is

$i'(x)=\exp(a^{x}\ln x)[a^{x}\ln a\ln x+a^{x}\cdot {\tfrac {1}{x}}]=x^{(a^{x})}a^{x}[\ln a\ln x+{\tfrac {1}{x}})]$

Part 5: There is $j(x)=a^{(x^{x})}=\exp(x^{x}\ln a)$. The function $x\mapsto x^{x}=\exp(x\ln(x))$ is differentiable with $(x^{x})'=x^{x}(\ln(x)+1)$. Hence $j$ is differentiable by the chain- and product rule and for $x\in \mathbb {R} ^{+}$ there is

$j'(x)=\exp(x^{x}\ln a)[x^{x}(\ln x+1)\cdot \ln a]=a^{(x^{x})}x^{x}\ln a(\ln x+1)$

Exercise (Proof of sum formulas using the derivative)

Proofs by means of binomial theorem (missing) that for all $n\in \mathbb {N}$:

• $\sum _{k=1}^{n}k{\binom {n}{k}}=n2^{n-1}$
• $\sum _{k=1}^{n}(-1)^{k-1}k{\binom {n}{k}}=0$

Use the binomial theorem and set $x=1$ . Then take the derivative on both sides.

Proof (Proof of sum formulas using the derivative)

For $x=1$ the binomial theorem reads

$\sum _{k=0}^{n}{\binom {n}{k}}y^{k}=(1+y)^{n}$

for $n\in \mathbb {N}$ and $y\in \mathbb {R}$. Now the left-hand side of the equation is a polynomial $f(y)$ and the right-hand side is a power function $g(y)$. Both sides are therefore differentiable on $\mathbb {R}$ with

$f'(y)=\sum _{k=0}^{n}{\binom {n}{k}}ky^{k-1}=\sum _{k=1}^{n}k{\binom {n}{k}}y^{k-1}$

and

$g'(y)=n(1+y)^{n-1}$

Since $f\equiv g$ there is also $f'\equiv g'$. So in particular

$f(1)=g(1)\iff \sum _{k=1}^{n}k{\binom {n}{k}}=n2^{n-1}$

and

$f(-1)=g(-1)\iff \sum _{k=1}^{n}(-1)^{k-1}k{\binom {n}{k}}=0$

Exercise (Logarithmic derivatives)

Determine the logarithmic derivatives of the following functions

1. $f(x)={\tfrac {1}{x}}$
2. $g(x)=\sec(x)={\tfrac {1}{\sin(x)}}$
3. $h(x)=a^{x}=\exp(x\ln(a))$ with $a\in \mathbb {R} ^{+}$
4. $h(x)=x^{x}=\exp(x\ln(x))$

Proof of computational laws

Exercise (Alternative proof of the product rule)

Prove that for differentiable $f,g:D\to \mathbb {R}$ the product rule

$(fg)'=f'g+g'f$

holds - by using the chain rule.

Hint: There is: $fg={\tfrac {1}{2}}((f+g)^{2}-f^{2}-g^{2})$

Proof (Alternative proof of the product rule)

The function $g:\mathbb {R} \to \mathbb {R} ,\ g(x)=x^{2}$ is differentiable on $\mathbb {R}$ with

$g'(x)=2x$

By der chain rule, we hence have that $g\circ f=f^{2}:D\to \mathbb {R}$ is differentiable with

$(g\circ f)'(x)=(g\circ f)'(x)\cdot f'(x)=2f(x)f'(x)$

for all $x\in D$. Using the hint, we get with the factor- and sum rule

${\begin{aligned}(fg)'(x)&=({\tfrac {1}{2}}((f+g)^{2}-f^{2}-g^{2}))'(x)\\[0.3em]&={\tfrac {1}{2}}[(2(f(x)+g(x)))(f'(x)+g'(x))-2f(x)f'(x)-2g(x)g'(x)]\\[0.3em]&={\tfrac {1}{2}}[2f(x)f'(x)+2f(x)g'(x)+2g(x)f'(x)+2g(x)g'(x)-2f(x)f'(x)-2g(x)g'(x)]\\[0.3em]&={\tfrac {1}{2}}[2f(x)g'(x)+2g(x)f'(x)]\\[0.3em]&=f(x)g'(x)+g(x)f'(x)\end{aligned}}$

Exercise (Special case of the chain rule)

Derive a general derivative formula for the following function:

$f_{1}^{f_{2}}:(0,\infty )\to \mathbb {R} ,\ (f_{1}^{f_{2}})(x)=\exp(f_{2}(x)\ln(f_{1}(x)))$

If $f_{1},f_{2}:(0,\infty )\to \mathbb {R}$ are differentiable.

Solution (Special case of the chain rule)

There is

$f_{1}^{f_{2}}:(0,\infty )\to \mathbb {R} ,\ f_{1}(x)^{f_{2}(x)}=\exp(f_{2}(x)\cdot \ln(f_{1}(x)))=g(f(x))$

with $g(x)=\exp(x)$ and $f(x)=f_{2}(x)\cdot \ln(f_{1}(x))$ for all $x\in (0,\infty )$. The function $f$ is differentiable by the product rule with

$f'(x)=f_{2}'(x)\ln(f_{1}(x))+f_{2}(x){\tfrac {f_{1}'(x)}{f_{1}(x)}}$

By the chain rule, also $f_{1}^{f_{2}}:(0,\infty )\to \mathbb {R}$ is differentiable, and there is

$(f_{1}^{f_{2}})'(x)=g'(f(x))\cdot f'(x)=\exp(f_{2}(x))\cdot [f_{2}'(x)\ln(f_{1}(x))+f_{2}(x){\tfrac {f_{1}'(x)}{f_{1}(x)}}]=f_{1}(x)^{f_{2}(x)}[f_{2}'(x)\ln(f_{1}(x))+f_{2}(x){\tfrac {f_{1}'(x)}{f_{1}(x)}}]$

Theorem (Computational laws for logarithmic derivatives)

For two differentiable functions $f$ and $g$ without zeros there is

1. $L({\sqrt[{k}]{f}})={\tfrac {1}{k}}L(f)$ for $k\in \mathbb {N}$ and $f>0$
2. $L(f^{\alpha })=\alpha L(f)$ for $\alpha \in \mathbb {R}$ and $f>0$
3. $L(f+g)=L(f)+L(1+{\tfrac {f}{g}})$