Examples for derivatives – Serlo

In this chapter we want to summarise the most important examples of derivatives. The derivative rules will allow us for computing derivatives of composite functions.

Table of important derivatives

In the following table $n\in \mathbb {N}$, $q\in \mathbb {Z}$ and ${\tilde {n}}\in \mathbb {N} _{0}$ is given. We also define $a,b,c\in \mathbb {R}$, $a_{k}\in \mathbb {R}$ and $p\in \mathbb {R} ^{+}$.

function term	term of the derivative function	domain of definition of the derivative
$c$	$0$	$\mathbb {R}$
$x^{n}$	$nx^{n-1}$	$\mathbb {R}$
$ax+b$	$a$	$\mathbb {R}$
$ax^{2}+bx+c$	$2ax+b$	$\mathbb {R}$
$\sum _{k=0}^{\tilde {n}}a_{k}x^{k}$	$\sum _{k=1}^{\tilde {n}}a_{k}kx^{k-1}$	$\mathbb {R}$
${\frac {1}{x}}$	$-{\frac {1}{x^{2}}}$	$\mathbb {R} \setminus \{0\}$
$x^{-n}={\frac {1}{x^{n}}}$	$-{\frac {n}{x^{n+1}}}$	$\mathbb {R} \setminus \{0\}$
$x^{q}$	$qx^{q-1}$	${\begin{cases}\mathbb {R} &q\in \mathbb {Z} _{0}^{+}\\\mathbb {R} \setminus \{0\}&q\in \mathbb {Z} ^{-}\end{cases}}$
${\sqrt {x}}$	${\frac {1}{2{\sqrt {x}}}}$	$\mathbb {R} ^{+}$
${\sqrt[{n}]{x}}$	${\frac {1}{n{\sqrt[{n}]{x^{n-1}}}}}$	$\mathbb {R} ^{+}$
${\sqrt[{n}]{x^{q}}}$	${\frac {q}{n}}{\sqrt[{n}]{x^{q-n}}}$	$\mathbb {R} ^{+}$
$\exp(x)=e^{x}$	$\exp(x)$	$\mathbb {R}$
$p^{x}=\exp(x\ln p)$	$\ln(p)\cdot p^{x}$	$\mathbb {R}$
$x^{a}=\exp(a\ln x)$	$ax^{a-1}$	${\begin{cases}\mathbb {R} &a\geq 0\\\mathbb {R} \setminus \{0\}&a<0\end{cases}}$
$\ln |x|$	${\frac {1}{x}}$	$\mathbb {R} \setminus \{0\}$
$\log _{p}|x|={\frac {\ln |x|}{\ln p}}$	${\frac {1}{\ln(p)\cdot x}}$	$\mathbb {R} \setminus \{0\}$
$\sin(x)$	$\cos(x)$	$\mathbb {R}$
$\cos(x)$	$-\sin(x)$	$\mathbb {R}$
$\tan(x)={\frac {\sin x}{\cos x}}$	${\frac {1}{\cos ^{2}(x)}}=1+\tan ^{2}(x)$	$\mathbb {R} \setminus \{{\frac {\pi }{2}}+k\pi |k\in \mathbb {Z} \}$
$\sec(x)={\frac {1}{\cos(x)}}$	${\frac {\sin(x)}{\cos ^{2}(x)}}$	$\mathbb {R} \setminus \{{\frac {\pi }{2}}+k\pi |k\in \mathbb {Z} \}$
$\csc(x)={\frac {1}{\sin(x)}}$	$-{\frac {\cos(x)}{\sin ^{2}(x)}}$	$\mathbb {R} \setminus \{k\pi |k\in \mathbb {Z} \}$
$\cot(x)={\frac {\cos x}{\sin x}}$	$-{\frac {1}{\sin ^{2}(x)}}=-1-\cot ^{2}(x)$	$\mathbb {R} \setminus \{k\pi |k\in \mathbb {Z} \}$
$\arcsin(x)$	${\frac {1}{\sqrt {1-x^{2}}}}$	$(-1,1)$
$\arccos(x)$	$-{\frac {1}{\sqrt {1-x^{2}}}}$	$(-1,1)$
$\arctan(x)$	${\frac {1}{1+x^{2}}}$	$\mathbb {R}$
${\text{arcot}}(x)$	$-{\frac {1}{1+x^{2}}}$	$\mathbb {R}$
$\sinh(x)={\frac {e^{x}-e^{-x}}{2}}$	$\cosh(x)$	$\mathbb {R}$
$\cosh(x)={\frac {e^{x}+e^{-x}}{2}}$	$\sinh(x)$	$\mathbb {R}$
$\tanh(x)={\frac {\sinh x}{\cosh x}}$	${\frac {1}{\cosh ^{2}(x)}}=1-\tanh ^{2}(x)$	$\mathbb {R}$
$\operatorname {arsinh} (x)$	${\frac {1}{\sqrt {x^{2}+1}}}$	$\mathbb {R}$
$\operatorname {arcosh} (x)$	${\frac {1}{\sqrt {x^{2}-1}}}$	$(1,\infty )$
$\operatorname {artanh} (x)$	${\frac {1}{1-x^{2}}}$	$(-1,1)$

Examples for computing derivatives

Now we will calculate some examples of derivatives from the table above. Often it comes down to determining the differential quotient of the function, i.e. a limit value. But sometimes it is also useful to use the calculation rules from the chapter before.

Constant functions

We start with some simple derivatives:

Theorem (Derivative of a constant function)

Every constant function $f\equiv c$ is differentiable on all of $\mathbb {R}$ with derivative $0$.

Proof (Derivative of a constant function)

Let ${\tilde {x}}\in \mathbb {R}$. Then there is

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

Power functions with natural numbers as powers

Now we turn to the derivative of power functions with natural powers. First we will deal with a few special cases:

Example (Derivative of the identity and the square function)

The functions

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

and

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

are differentiable on all of $\mathbb {R}$. Further there is for ${\tilde {x}}\in \mathbb {R}$:

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

as well as

$g'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{2}-{\tilde {x}}^{2}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {(x+{\tilde {x}})(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}(x+{\tilde {x}})=2{\tilde {x}}$

For the derivative of $g$ we used the 3rd binomial formula $(x+y)(x-y)=x^{2}-y^{2}$.

Exercise (Derivative of a power function)

Compute die derivative von

$h:\mathbb {R} \to \mathbb {R} ,\ h(x)=x^{3}$

Solution (Derivative of a power function)

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

$h'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{3}-{\tilde {x}}^{3}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})(x-{\tilde {x}})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})=3{\tilde {x}}^{2}$

Instead of using the identity $x^{3}-{\tilde {x}}^{3}=(x^{2}+x{\tilde {x}}+{\tilde {x}}^{2})(x-{\tilde {x}})$, we could also have calculated ${\tfrac {x^{3}-{\tilde {x}}^{3}}{x-{\tilde {x}}}}=x^{2}+x{\tilde {x}}+{\tilde {x}}^{2}$ using polynomial division.

Now we turn to the general case, i.e. the derivative of $x\mapsto x^{n}$ for $n\in \mathbb {N}$:

Theorem (Derivative of power functions)

The power function

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

is for $n\in \mathbb {N}$ differentiable on all of $\mathbb {R}$. For alle ${\tilde {x}}\in \mathbb {R}$ there is

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

Proof (Derivative of power functions)

For ${\tilde {x}}\in \mathbb {R}$, so there is

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {x^{n}-{\tilde {x}}^{n}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}\sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}=\sum _{k=0}^{n-1}{\tilde {x}}^{k}{\tilde {x}}^{n-1-k}=\sum _{k=0}^{n-1}{\tilde {x}}^{n-1}=n{\tilde {x}}^{n-1}$

We used the geometric sum formula $\sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}={\tfrac {x^{n}-{\tilde {x}}^{n}}{x-{\tilde {x}}}}$ and the continuity of the polynomial function $x\mapsto \sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}$.

Polynomials and rational functions

Using the calculation rules for derivatives we can now calculate the derivatives of polynomial functions and rational functions:

Theorem (Derivative of polynomial functions)

Let

$p:\mathbb {R} \to \mathbb {R} ,\ p(x)=\sum _{k=0}^{n}a_{k}x^{k}$

with $a_{k}\in \mathbb {R}$ and $n\in \mathbb {N}$ be a polynomial function of degree $n$. Then $p$ is differentiable on all of $\mathbb {R}$ , and for ${\tilde {x}}\in \mathbb {R}$ there is

$p'({\tilde {x}})=\sum _{k=1}^{n}a_{k}k{\tilde {x}}^{k-1}$

Proof (Derivative of polynomial functions)

Using the derivative rule for the multiple of a function $(\lambda f)'=\lambda f'$ , every single summand of the polynomial is differentiable on $\mathbb {R}$. With the summation rule we can derive every polynomial function term by term on $\mathbb {R}$ and obtain for ${\tilde {x}}\in \mathbb {R}$:

$p'({\tilde {x}})=\sum _{k=1}^{n}a_{k}k{\tilde {x}}^{k-1}$

with the derivative of the zeroth summand disappearing.

In particular, it follows for $n=1$ and $n=2$ that linear and quadratic functions are differentiable onn all of $\mathbb {R}$.

Exercise (Derivative of rational functions)

Let

$r(x)={\frac {\sum _{k=0}^{n}a_{k}x^{k}}{\sum _{l=0}^{m}b_{l}x^{l}}}$

with $a_{k},b_{l}\in \mathbb {R}$ and $n,m\in \mathbb {N} _{0}$ a rational function defined on $D=\mathbb {R} \setminus \{x\in \mathbb {R} :\sum _{l=0}^{m}b_{l}x^{l}\neq 0\}$. Show that $r$ is differentiable on $D$, and calculate the derivative.

Solution (Derivative of rational functions)

Numerator and denominator of $r$ are polynomials. Since the denominator is non-zero on $D$ and polynomials are differentiable, it follows from the quotient rule that $r$ is differentiable on $D$.

Further there is for ${\tilde {x}}\in D$:

$r'({\tilde {x}})={\frac {\left(\sum _{k=1}^{n}a_{k}k{\tilde {x}}^{k-1}\right)\cdot \left(\sum _{l=0}^{m}b_{l}{\tilde {x}}^{l}\right)-\left(\sum _{k=0}^{n}a_{k}{\tilde {x}}^{k}\right)\cdot \left(\sum _{l=1}^{m}b_{l}l{\tilde {x}}^{l-1}\right)}{\left(\sum _{l=0}^{m}b_{l}{\tilde {x}}^{l}\right)^{2}}}$

Power functions with integer powers

We can already differentiate power functions with natural powers. Now we investigate those with negative integer exponents.

Example (Derivative of the hyperbolic function)

The power function

$f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-1}={\frac {1}{x}}$

is differentiable on $\mathbb {R} \setminus \{0\}$ and there is

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x}}-{\frac {1}{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}-x}{x{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {{\tilde {x}}-x}{x{\tilde {x}}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x-{\tilde {x}})}{x{\tilde {x}}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-1}{x{\tilde {x}}}}=-{\frac {1}{{\tilde {x}}^{2}}}$

for ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$.

Exercise (Derivative of $x\mapsto {\tfrac {1}{x^{2}}}$)

Prove that the power function

$f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-2}={\frac {1}{x^{2}}}$

is differentiable on $\mathbb {R} \setminus \{0\}$ and compute its derivative.

Solution (Derivative of $x\mapsto {\tfrac {1}{x^{2}}}$)

For ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x^{2}}}-{\frac {1}{{\tilde {x}}^{2}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}^{2}-x^{2}}{x^{2}{\tilde {x}}^{2}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {({\tilde {x}}-x)({\tilde {x}}+x)}{x^{2}{\tilde {x}}^{2}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x-{\tilde {x}})(x+{\tilde {x}})}{x^{2}{\tilde {x}}^{2}(x-{\tilde {x}})}}=\lim _{x\to {\tilde {x}}}{\frac {-(x+{\tilde {x}})}{x^{2}{\tilde {x}}^{2}}}=-{\frac {2{\tilde {x}}}{{\tilde {x}}^{4}}}=-{\frac {2}{{\tilde {x}}^{3}}}$

In the general case $x^{-n}={\tfrac {1}{x^{n}}}$ with $n\in \mathbb {N}$ there is

Theorem (Derivative of the power function with negative integer powers)

The power function

$f:\mathbb {R} \setminus \{0\},\ f(x)=x^{-n}={\frac {1}{x^{n}}}$

is differentiable on $\mathbb {R} \setminus \{0\}$, and for ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is

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

Proof (Derivative of the power function with negative integer powers)

For ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is

${\begin{aligned}f'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {{\frac {1}{x^{n}}}-{\frac {1}{{\tilde {x}}^{n}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {\frac {{\tilde {x}}^{n}-x^{n}}{x^{n}{\tilde {x}}^{n}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}{\frac {{\tilde {x}}^{n}-x^{n}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}{\frac {-({\tilde {x}}^{n}-x^{n})}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{x^{n}{\tilde {x}}^{n}}}\cdot \left(-\sum _{k=0}^{n-1}x^{k}{\tilde {x}}^{n-1-k}\right)\\&={\frac {1}{{\tilde {x}}^{2n}}}\cdot \left(-\sum _{k=0}^{n-1}{\tilde {x}}^{n-1}\right)=-{\frac {1}{{\tilde {x}}^{2n}}}\cdot n{\tilde {x}}^{n-1}=-{\frac {n}{{\tilde {x}}^{n-1}}}\end{aligned}}$

Exercise (Derivative of the power function)

Prove $({\tfrac {1}{x^{n}}})'={\tfrac {-n}{x^{n+1}}}$ using the quotient rule

Solution (Derivative of the power function)

For ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is by the quotient rule

$\left({\frac {1}{{\tilde {x}}^{n}}}\right)'={\frac {0\cdot {\tilde {x}}^{n}-1\cdot n{\tilde {x}}^{n-1}}{({\tilde {x}}^{n})^{2}}}={\frac {-n{\tilde {x}}^{n-1}}{{\tilde {x}}^{2n}}}={\frac {-n}{{\tilde {x}}^{2n-(n-1)}}}={\frac {-n}{{\tilde {x}}^{n+1}}}$

Remark: Of course we can also apply the inverse rule directly, and thus get the same result

$\left({\frac {1}{{\tilde {x}}^{n}}}\right)'=-{\frac {n{\tilde {x}}^{n-1}}{{\tilde {x}}^{2n}}}={\frac {-n}{{\tilde {x}}^{n+1}}}$

Let us look again at the derivatives rule in the last case, i.e. $f'({\tilde {x}})=-{\tfrac {n}{{\tilde {x}}^{n-1}}}=-n{\tilde {x}}^{-n-1}$ for $n\in \mathbb {N}$. If we put $k=-n\in \mathbb {Z} ^{-}$, we get $f'({\tilde {x}})=k{\tilde {x}}^{k-1}$. The derivative rule is hence the same as for $x^{n}$ with $n\in \mathbb {N}$. So we can summarize the two cases and get

Theorem (Derivative of the power function with natural powers)

For $k\in \mathbb {Z}$ the power function

$f:\mathbb {R} \setminus \{0\},\ f(x)=x^{k}$

is differentiable on $\mathbb {R} \setminus \{0\}$. For ${\tilde {x}}\in \mathbb {R} \setminus \{0\}$ there is then

$f'({\tilde {x}})=kx^{k-1}$

In the case of $k\in \mathbb {N}$ it is even differentiable on all of $\mathbb {R}$.

Root functions

Now we investigate the derivative of root functions. We start again with the simplest case:

Example (Derivative of the square root function)

The square root function

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

is differentiable on $\mathbb {R} ^{+}$ and for ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

${\begin{aligned}f'({\tilde {x}})&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{x-{\tilde {x}}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{({\sqrt {x}})^{2}-({\sqrt {\tilde {x}}})^{2}}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt {x}}-{\sqrt {\tilde {x}}}}{({\sqrt {x}}-{\sqrt {\tilde {x}}})({\sqrt {x}}+{\sqrt {\tilde {x}}})}}\\[1em]&=\lim _{x\to {\tilde {x}}}{\frac {1}{{\sqrt {x}}+{\sqrt {\tilde {x}}}}}\\[1em]&={\frac {1}{2{\sqrt {\tilde {x}}}}}\end{aligned}}$

Question: Why is the square root function in ${\tilde {x}}=0$ not differentiable, although it is defined and continuous there?

For the differential quotient there is

$\lim _{x\to 0+}{\frac {{\sqrt {x}}-{\sqrt {0}}}{x-0}}=\lim _{x\to 0+}{\frac {\sqrt {x}}{x}}=\lim _{x\to 0+}{\frac {1}{\sqrt {x}}}=\infty$

So it does not exist. Hence, we have non-differentiability.

Exercise (Derivative of the cubic root function)

Compute the derivative of the cubic root function

$f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)={\sqrt[{3}]{x}}$

Solution (Derivative of the cubic root function)

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

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

Now let us consider the general case of the $k$-th root function. Here there is

Theorem (Derivative of the $k$-th root function)

Let $k\in \mathbb {N} ,k>1$. Then the $k$-th root function

$f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)={\sqrt[{k}]{x}}$

is differentiable on $\mathbb {R} ^{+}$, and for ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

$f'({\tilde {x}})={\frac {1}{k{\sqrt[{k}]{x^{k-1}}}}}$

Proof (Derivative of the $k$-th root function)

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

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}{x-{\tilde {x}}}}=\lim _{x\to {\tilde {x}}}{\frac {{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}{({\sqrt[{k}]{x}})^{k}-({\sqrt[{k}]{\tilde {x}}})^{k}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{\frac {({\sqrt[{k}]{x}})^{k}-({\sqrt[{k}]{\tilde {x}}})^{k}}{{\sqrt[{k}]{x}}-{\sqrt[{k}]{\tilde {x}}}}}}=\lim _{x\to {\tilde {x}}}{\frac {1}{\sum _{l=0}^{k-1}({\sqrt[{k}]{x}})^{l}({\sqrt[{k}]{\tilde {x}}})^{k-1-l}}}={\frac {1}{\sum _{l=0}^{k-1}({\sqrt[{k}]{\tilde {x}}})^{k-1}}}={\frac {1}{k({\sqrt[{k}]{\tilde {x}}})^{k-1}}}={\frac {1}{k{\sqrt[{k}]{{\tilde {x}}^{k-1}}}}}$

This can now be generalised

Theorem (Derivative of the generalized root function)

For $q\in \mathbb {N} ,q>1$ and $p\in \mathbb {Z}$, the generalized root function

$f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)={\sqrt[{q}]{x}}^{p}$

is differentiable on $\mathbb {R} ^{+}$, and for ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

$f'({\tilde {x}})={\frac {p}{q}}{\sqrt[{q}]{x^{p-q}}}$

Proof (Derivative of the generalized root function)

Since on $\mathbb {R} ^{+}$ die functions $x\mapsto x^{p}$ and $y\mapsto {\sqrt[{q}]{y}}$ are differentiable, the chain rule implies at ${\tilde {x}}\in \mathbb {R} ^{+}$ that

$f'({\tilde {x}})={\frac {1}{q{\sqrt[{q}]{({\tilde {x}}^{p})^{q-1}}}}}\cdot p{\tilde {x}}^{p-1}={\frac {p}{q}}{\frac {\sqrt[{q}]{({\tilde {x}}^{p-1})^{q}}}{\sqrt[{q}]{({\tilde {x}}^{p})^{q-1}}}}={\frac {p}{q}}{\sqrt[{q}]{\frac {{\tilde {x}}^{pq-q}}{{\tilde {x}}^{pq-p}}}}={\frac {p}{q}}{\sqrt[{q}]{{\tilde {x}}^{pq-q-pq+p}}}={\frac {p}{q}}{\sqrt[{q}]{{\tilde {x}}^{p-q}}}$

Hint

For $p\in \mathbb {Z}$ and $q\in \mathbb {N}$ and $r={\tfrac {p}{q}}\in \mathbb {Q}$ the power fucniton with rational exponent was defined as

$x^{r}=x^{\frac {p}{q}}={\sqrt[{q}]{x^{p}}}$

So for $r\in \mathbb {Q}$ we also have the derivative rule

$(x^{r})'={\frac {p}{q}}{\sqrt[{q}]{{\tilde {x}}^{p-q}}}={\frac {p}{q}}x^{\frac {p-q}{q}}={\frac {p}{q}}x^{{\frac {p}{q}}-1}=rx^{r-1}$

The (generalized) exponential function and generalized power functions

In this section we prove that the derivative of the exponential function is again the exponential function. So we can determine the derivative of the generalized exponential and power function.

Theorem (Derivative of the exponential function)

The exponential function

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

is differentiable on $\mathbb {R}$, and for ${\tilde {x}}\in \mathbb {R}$ there is

$f'({\tilde {x}})=\exp({\tilde {x}})$

How to get to the proof? (Derivative of the exponential function)

For this derivative it is more useful to use the $h$ method

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

Because in this case we know the limit value

$\lim _{h\to 0}{\frac {\exp(h)-1}{h}}=1$

Furthermore we need the functional equation of the exponential function

$\exp(x+y)=\exp(x)\exp(y)$

Proof (Derivative of the exponential function)

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

${\begin{aligned}f'({\tilde {x}})&=\lim _{h\to 0}{\frac {f({\tilde {x}}+h)-f({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\exp({\tilde {x}}+h)-\exp({\tilde {x}})}{h}}=\lim _{h\to 0}{\frac {\exp({\tilde {x}})\exp(h)-\exp({\tilde {x}})}{h}}\\&=\lim _{h\to 0}\exp({\tilde {x}}){\frac {\exp(h)-1}{h}}=\exp({\tilde {x}})\cdot \lim _{h\to 0}{\frac {\exp(h)-1}{h}}=\exp({\tilde {x}})\cdot 1=\exp({\tilde {x}})\end{aligned}}$

Using the chain rule, the derivatives of the generalized exponential function $x\mapsto a^{x}$ for $a\in \mathbb {R} ^{+}$ and the generalized power function $x\mapsto x^{r}$ for $r\in \mathbb {R}$ can be calculated:

Theorem (Derivative of the generalized exponential function)

For $a\in \mathbb {R} ^{+}$ the generalized exponential function

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

is differentiable on $\mathbb {R}$, and for ${\tilde {x}}\in \mathbb {R}$ there is

$f'({\tilde {x}})=\ln(a)a^{\tilde {x}}$

Proof (Derivative of the generalized exponential function)

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

$f'({\tilde {x}})=\exp({\tilde {x}}\ln(a))\cdot \ln(a)=\ln(a)a^{\tilde {x}}$

Theorem (Derivative of the generalized exponential function)

For $r\in \mathbb {R}$ the generalized exponential function

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

is differentiable on $\mathbb {R} ^{+}$, and for ${\tilde {x}}\in \mathbb {R} ^{+}$ there is

$f'({\tilde {x}})=r{\tilde {x}}^{r-1}$

Exercise (Derivative of the generalized exponential function)

Prove that the derivative of the generalized power function at ${\tilde {x}}\in \mathbb {R} ^{+}$ is $r{\tilde {x}}^{r-1}$.

Proof (Derivative of the generalized exponential function)

For ${\tilde {x}}\in \mathbb {R}$ the chain rule yields

$f'({\tilde {x}})=\exp(\ln({\tilde {x}})r)\cdot {\frac {r}{\tilde {x}}}=r{\tilde {x}}^{r}{\frac {1}{\tilde {x}}}=r{\tilde {x}}^{r-1}$

Logarithmic functions

Now we turn to the derivative of the natural and generalised logarithm function. Since the natural logarithm is the inverse of the exponential function, we can deduce its derivative directly from rule for derivatives of inverse function:

Theorem (Derivative of the natural logarithm function)

The natural logarithm function

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

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

$g'(y)={\frac {1}{y}}$

Proof (Derivative of the natural logarithm function)

For the exponential function $\exp :\mathbb {R} \to \mathbb {R}$ there is: $f'=\exp$. So the function is differentiable, and because of $f'>0$ strictly monotonously increasing. Furthermore, $f$ is surjective. The inverse function $f^{-1}=g$ is the (natural) logarithm function

$f^{-1}:\mathbb {R} ^{+}\to \mathbb {R} ,\ f^{-1}(x)=\ln x$

From the theorem about the derivative of the inverse function we now have for every $y\in \mathbb {R} ^{+}$:

$g'(y)=(f^{-1})'(y)={\frac {1}{f'(f^{-1}(y))}}={\frac {1}{e^{\ln y}}}={\frac {1}{y}}$

The derivative can also be calculated directly using the differential quotient. If you want to try this, we recommend the corresponding exercise (missing).

Using the derivative of the natural logarithm function we can now immediately conclude

Theorem (Derivative of the generalized logarithm function)

For $a\in \mathbb {R} ^{+}$ the generalized logarithm function

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

is differentiable on $\mathbb {R} ^{+}$. For $y\in \mathbb {R} ^{+}$ there is

$g'(y)={\frac {1}{y\ln a}}$

Proof (Derivative of the generalized logarithm function)

From the derivative rule for the multiple of a function, we get that for all $y\in \mathbb {R} ^{+}$:

$g'(y)={\frac {1}{\ln a}}{\frac {1}{y}}={\frac {1}{y\ln a}}$

If the derivative of the natural logarithm is not available, we can calculate it using the theorem of the derivative of the inverse function.

Trigonometric functions

Sine

Theorem (Derivative of the sine function)

The sine function is differentiable. For all ${\tilde {x}}\in \mathbb {R}$ there is:

$\sin '({\tilde {x}})=\cos({\tilde {x}})$

Proof (Derivative of the sine function)

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

${\begin{aligned}\sin '({\tilde {x}})&=\lim _{h\to 0}{\frac {\sin({\tilde {x}}+h)-\sin({\tilde {x}})}{h}}\\[0.3em]&\color {Gray}\left\downarrow \ \sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\right.\\[0.3em]&=\lim _{h\to 0}{\frac {\sin({\tilde {x}})\cos(h)+\cos({\tilde {x}})\sin(h)-\sin({\tilde {x}})}{h}}\\[0.3em]&=\sin({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\cos(h)-1}{h}}} _{=0}+\cos({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\sin(h)}{h}}} _{=1}\\[0.3em]&=\cos({\tilde {x}})\end{aligned}}$

Cosine

Theorem (Derivative of the cosine function)

The cosine function is differentiable with

$\cos '({\tilde {x}})=-\sin({\tilde {x}})$

Proof (Derivative of the cosine function)

${\begin{aligned}\cos '({\tilde {x}})&=\lim _{h\to 0}{\frac {\cos({\tilde {x}}+h)-\cos({\tilde {x}})}{h}}\\[0.5em]&\color {Gray}\left\downarrow \ \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\right.\\[0.5em]&=\lim _{h\to 0}{\frac {\cos({\tilde {x}})\cos(h)-\sin({\tilde {x}})\sin(h)-\cos({\tilde {x}})}{h}}\\[0.5em]&=\cos({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\cos(h)-1}{h}}} _{=0}-\sin({\tilde {x}})\cdot \underbrace {\lim _{h\to 0}{\frac {\sin(h)}{h}}} _{=1}\\[0.5em]&=-\sin({\tilde {x}})\end{aligned}}$

Tangent

Theorem (Derivative of the tangent function)

The tangent function

$\tan :D=\mathbb {R} \setminus \{{\tfrac {\pi }{2}}+k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \tan(x)={\frac {\sin(x)}{\cos(x)}}$

is differentiable on $D$, and for ${\tilde {x}}\in D$ there is

$\tan '({\tilde {x}})={\frac {1}{\cos ^{2}({\tilde {x}})}}=1+\tan ^{2}({\tilde {x}})$

Proof (Derivative of the tangent function)

Since $\cos(x)\neq 0$ for $x\neq {\tfrac {\pi }{2}}+k\pi$ , the function $\tan$ is differentiable by the quotient rule, and for ${\tilde {x}}\in D$ there is

$\tan '({\tilde {x}})={\frac {\cos({\tilde {x}})\cos({\tilde {x}})-\sin({\tilde {x}})(-\sin({\tilde {x}}))}{\cos ^{2}({\tilde {x}})}}={\frac {\cos ^{2}({\tilde {x}})+\sin ^{2}({\tilde {x}})}{\cos ^{2}({\tilde {x}})}}{\begin{cases}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}{\frac {1}{\cos ^{2}({\tilde {x}})}}\\=1+{\frac {\sin ^{2}({\tilde {x}})}{\cos ^{2}({\tilde {x}})}}=1+\tan ^{2}({\tilde {x}})\end{cases}}$

Exercise (Derivative of the cotangent function)

The cotangent function

$\cot :D=\mathbb {R} \setminus \{k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \cot(x)={\frac {1}{\tan(x)}}={\frac {\cos(x)}{\sin(x)}}$

is differentiable on $D$, and for ${\tilde {x}}\in D$ there is

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

Solution (Derivative of the cotangent function)

Since $\sin(x)\neq 0$ for $x\neq k\pi$ , the function $\tan$ is differentiable by the quotient rule, and for ${\tilde {x}}\in D$ there is

$\cot '({\tilde {x}})=\left({\frac {\cos(x)}{\sin(x)}}\right)'={\frac {-\sin({\tilde {x}})\sin({\tilde {x}})-\cos({\tilde {x}})\cos({\tilde {x}})}{\sin ^{2}({\tilde {x}})}}=-{\frac {\sin ^{2}({\tilde {x}})+\cos ^{2}({\tilde {x}})}{\sin ^{2}({\tilde {x}})}}{\begin{cases}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}-{\frac {1}{\sin ^{2}({\tilde {x}})}}\\=-1-{\frac {\cos ^{2}({\tilde {x}})}{\sin ^{2}({\tilde {x}})}}=-1-\cot ^{2}({\tilde {x}})\end{cases}}$

Alternative solution:

$\cot '({\tilde {x}})=\left({\frac {1}{\tan(x)}}\right)'=-{\frac {1+\tan ^{2}({\tilde {x}})}{\tan ^{2}({\tilde {x}})}}{\begin{cases}=-{\frac {1+{\frac {\sin ^{2}({\tilde {x}})}{\cos ^{2}({\tilde {x}})}}}{\frac {\sin ^{2}({\tilde {x}})}{\cos ^{2}({\tilde {x}})}}}=-{\frac {\cos ^{2}({\tilde {x}})+\sin ^{2}({\tilde {x}})}{\sin ^{2}({\tilde {x}})}}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}-{\frac {1}{\sin ^{2}({\tilde {x}})}}\\=-{\frac {1}{\tan ^{2}({\tilde {x}})}}-1=-1-\cot ^{2}({\tilde {x}})\end{cases}}$

The derivatives of secant and cosecant can be found in the corresponding exercise.

arc-functions

Using the rule for derivatives of the inverse function we can differentiate the arc-functions (which are inverses of sine, cosine, etc.)

arcsin and arccos

Theorem (Derivative of the arcsin/arccos function)

The inverse functions of the trigonometric functions $\arcsin$, $\arccos$ are differentiable with

${\begin{aligned}\arcsin '({\tilde {x}})&={\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}\quad {\text{ for all }}{\tilde {x}}\in (-1,1),\\\arccos '({\tilde {x}})&=-{\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}\quad {\text{ for all }}{\tilde {x}}\in (-1,1),\\\end{aligned}}$

Note: $\arcsin$ and $\arccos$ are defined and continuous on $[-1,1]$ , but only differentiable on $(-1,1)$.

Proof (Derivative of the arcsin/arccos function)

Derivative of $\arcsin$:

For the sine function $\sin :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R}$ there is: $\sin '=\cos$. So the function is differentiable, and since $\cos(x)>0$ for all $x\in (-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})$, it is strictly monotonously increasing on this interval. Further, $\sin((-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}))=(-1,1)$. So $\sin :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to (-1,1)$ is surjective. The inverse function is the arc sine function

$\arcsin :(-1,1)\to (-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R}$

From the theorem about the derivative of the inverse we now have for every ${\tilde {x}}\in (-1,1)$:

$\arcsin '({\tilde {x}})={\frac {1}{\sin '(\arcsin({\tilde {x}}))}}={\frac {1}{\cos(\arcsin({\tilde {x}}))}}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}{\frac {1}{\sqrt {1-\sin ^{2}(\arcsin({\tilde {x}}))}}}={\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}$

Derivative of $\arccos$:

For the cosine function $\cos :(0.\pi )\to \mathbb {R}$ there is: $\cos '=-\sin$. So the function is differentiable, and because of $-\sin(x)|_{(0,\pi )}<0$, strictly monotonously decreasing. Further, $\cos((0,\pi ))=(-1,1)$. So $\cos :(0,\pi )\to (-1,1)$ is surjective. The inverse function

$\arccos :(-1,1)\to (0,\pi )\to \mathbb {R}$

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

$\arccos '({\tilde {x}})={\frac {1}{\cos '(\arccos({\tilde {x}}))}}={\frac {1}{-\sin(\arcsin({\tilde {x}}))}}{\underset {\text{Pythagoras}}{\overset {\text{trigonometric}}{=}}}-{\frac {1}{\sqrt {1-\cos ^{2}(\arccos({\tilde {x}}))}}}=-{\frac {1}{\sqrt {1-{\tilde {x}}^{2}}}}$

arctan and arccot

Theorem (Derivative of the arctan/ arccot function)

The inverse functions of the trigonometric functions $\arctan$, $\operatorname {arccot}$ are differentiable, and there is

${\begin{aligned}\arctan '({\tilde {x}})&={\frac {1}{1+{\tilde {x}}^{2}}}\quad {\text{ for all }}{\tilde {x}}\in \mathbb {R} \\\operatorname {arccot} '({\tilde {x}})&=-{\frac {1}{1+{\tilde {x}}^{2}}}\quad {\text{ for all }}{\tilde {x}}\in \mathbb {R} \end{aligned}}$

Proof (Derivative of the arctan/ arccot function)

For the tangent function $\tan |_{(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})}$ there is: $\tan '=1+\tan ^{2}>0$. So the function is differentiable and strictly monotonically increasing. Further, $\tan((-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}))=\mathbb {R}$. So $\tan :(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}})\to \mathbb {R}$ is surjective. The inverse function

$\arccos :(-1,1)\to (0,\pi )\to \mathbb {R}$

is hence differentiable, and now for ${\tilde {x}}\in (-1,1)$ there is:

$\arctan '({\tilde {x}})={\frac {1}{\tan '(\arctan({\tilde {x}}))}}={\frac {1}{1+\tan ^{2}(\arctan({\tilde {x}}))}}={\frac {1}{1+{\tilde {x}}^{2}}}$

Hyperbolic functions

And finally, we determine the derivatives of the hyperbolic functions $\sinh$, $\cosh$ and $\tanh$:

Theorem (Derivative of hyperbolic functions)

The functions

${\begin{aligned}\sinh :\mathbb {R} \to \mathbb {R} ,\ \sinh(x)={\frac {e^{x}-e^{-x}}{2}}\\\cosh :\mathbb {R} \to \mathbb {R} ,\ \cosh(x)={\frac {e^{x}+e^{-x}}{2}}\\\tanh :\mathbb {R} \to \mathbb {R} ,\ \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}\end{aligned}}$

are differentiable, and there is

${\begin{aligned}\sinh '(x)&=\cosh(x)\\\cosh '({\tilde {x}})&=\sinh({\tilde {x}})\\\tanh '({\tilde {x}})&={\frac {1}{\cosh ^{2}({\tilde {x}})}}=\tanh ^{2}({\tilde {x}})-1\\\end{aligned}}$

Proof (Derivative of hyperbolic functions)

The derivatives follow directly from the calculation rules. We show only the derivative of $\sinh$. The other two are left to you for practice.

According to the factor and difference rule $\sinh(x)={\tfrac {1}{2}}e^{x}-{\tfrac {1}{2}}e^{-x}$ for all $x\in \mathbb {R}$ is differentiable, and there is

$\sinh '(x)={\tfrac {1}{2}}e^{x}-{\tfrac {1}{2}}e^{-x}(-1)={\tfrac {1}{2}}e^{x}+{\tfrac {1}{2}}e^{-x}=\cosh(x)$

Exercise (Derivative of $\cosh$ and $\tanh$)

Prove that $\cosh$ and $\tanh$ are differentiable with

$\cosh '=\sinh$ and $\tanh '={\frac {1}{\cosh }}=\tanh ^{2}-1$

Proof (Derivative of $\cosh$ and $\tanh$)

Derivative of $\cosh$:

According to the factor and sum rule, $\cosh(x)={\tfrac {1}{2}}e^{x}+{\tfrac {1}{2}}e^{-x}$ is differentiable for all $x\in \mathbb {R}$ , and there is

$\cosh '(x)={\tfrac {1}{2}}e^{x}+{\tfrac {1}{2}}e^{-x}(-1)={\tfrac {1}{2}}e^{x}-{\tfrac {1}{2}}e^{-x}=\sinh(x)$

Derivative of $\tanh$:

$\tanh ={\tfrac {\sinh }{\cosh }}$ is differentiable on all of $\mathbb {R}$ by the quotient rule and there is

$\tanh '(x)={\frac {\cosh(x)\cosh(x)-\sinh(x)\sinh(x)}{\cosh ^{2}(x)}}={\frac {\cosh ^{2}(x)-\sinh ^{2}(x)}{\cosh ^{2}(x)}}{\begin{cases}={\frac {1}{\cosh ^{2}(x)}}\\=1-{\frac {\sinh ^{2}(x)}{\cosh ^{2}(x)}}=1-\tanh ^{2}(x)\end{cases}}$