Power rule

The power rule is one of the basic rules of differential calculus in mathematics . It is used to determine the derivation of power functions .

Definition and scope

The derivative of the function is . This applies to and . ${\displaystyle f(x)=x^{n}}$ ${\displaystyle f'(x)=n\cdot x^{n-1}}$ ${\displaystyle n\in \{2,3,4,\ldots {}\}}$ ${\displaystyle x\in \mathbb {R} }$

For example, the function has the derivative . ${\displaystyle f(x)=x^{4}}$ ${\displaystyle f'(x)=4\cdot x^{3}}$

The power rule applies to only for , since the expression would appear at position 0 , the definition of which is not unique. ${\displaystyle n=1}$ ${\displaystyle x\in \mathbb {R} \setminus \{0\}}$ ${\displaystyle 0^{0}}$

The power rule applies to only for , otherwise a division by 0 would occur. ${\displaystyle n\in \{-1,-2,-3,\ldots {}\}}$ ${\displaystyle x\in \mathbb {R} \setminus \{0\}}$

The power rule also applies to power functions if the exponent ( exponent ) is not an integer, but only in the range : ${\displaystyle f(x)=x^{s}}$ ${\displaystyle s\in \mathbb {R} }$ ${\displaystyle x\in \mathbb {R} ,x>0}$

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

Derivation

1st case: The exponent is a natural number

The derivation of a power function at the point x is the limit value :

${\displaystyle (x^{n})'=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{n}-x^{n}}{\Delta x}}}$.

According to the binomial theorem , this is the same

${\displaystyle \lim _{\Delta x\to 0}{\frac {{n \choose 0}x^{n}+{n \choose 1}x^{n-1}{\Delta x}+{n \choose 2}x^{n-2}{\Delta x}^{2}+\dots +{n \choose n-1}x{\Delta x}^{n-1}+{n \choose n}{\Delta x}^{n}-x^{n}}{\Delta x}}}$

written with so-called binomial coefficients . The power rule follows from this:

${\displaystyle (x^{n})'=\lim _{\Delta x\to 0}{\frac {{n \choose 1}x^{n-1}{\Delta x}+{n \choose 2}x^{n-2}{\Delta x}^{2}+\dots +{n \choose n-1}x{\Delta x}^{n-1}+{n \choose n}{\Delta x}^{n}}{\Delta x}}}$

${\displaystyle =\lim _{\Delta x\to 0}\left[{n \choose 1}x^{n-1}+{n \choose 2}x^{n-2}{\Delta x}+\dots +{n \choose n-1}x{\Delta x}^{n-2}+{n \choose n}{\Delta x}^{n-1}\right]}$

${\displaystyle ={n \choose 1}x^{n-1}=n\cdot x^{n-1}}$.

Illustrated in the figure, an 'n-dimensional cube' grows in exactly n directions (along the n coordinate axes) by '(n-1) -dimensional cubes'. A square grows (or crystallizes) marginally by 2 side lines, and a cube grows by 3 squares.

2nd case: any exponent

One uses the representation with the help of the exponential function : and derives with the help of the chain rule and the derivation rule for the exponential function: ${\displaystyle x^{s}=(e^{\ln x})^{s}=e^{s\cdot \ln x}}$

${\displaystyle (x^{s})'=(e^{s\cdot \ln x})'=e^{s\cdot \ln x}\cdot (s\cdot \ln x)'}$

The factor rule and the rule for the derivation of the logarithm function are used for the inner derivative :

${\displaystyle (s\cdot \ln x)'=s\cdot {\frac {1}{x}}}$

By putting this in and writing for again , you get ${\displaystyle (e^{s\cdot \ln x})}$ ${\displaystyle x^{s}}$

${\displaystyle (x^{s})'=x^{s}\cdot s\cdot {\frac {1}{x}}=s\cdot x^{s-1}}$

This derivation only applies to . For, however , the function can also be differentiated at the point and the rule also applies to the point . One calculates directly using the difference quotient: ${\displaystyle x\neq 0}$ ${\displaystyle s>1}$ ${\displaystyle f(x)=x^{s}}$ ${\displaystyle x=0}$ ${\displaystyle x=0}$

${\displaystyle f'(0)=\lim _{\Delta x\to 0}{\frac {(\Delta x)^{s}-0^{s}}{\Delta x-0}}=\lim _{\Delta x\to 0}(\Delta x)^{s-1}=0=s\cdot 0^{s-1}}$

Multiple derivative of a power function with natural exponent

(For the notation of the following see Leibniz notation .) Within the domain of definition of a power function with a natural exponent , its -fold derivative is ... ${\displaystyle n}$ ${\displaystyle k}$

• ... for . ${\displaystyle 1\leq k\leq n:\quad {\frac {d^{k}}{dx^{k}}}x^{n}={\frac {n!}{(n-k)!}}\cdot x^{n-k}}$

proof

The claim can be proved for with complete induction . ${\displaystyle k\leq n}$ Induction start for (true) ${\displaystyle k=1:\quad {\frac {d}{dx}}x^{n}=n\cdot x^{n-1}={\frac {n!}{(n-1)!}}x^{n-1}}$ Induction requirement: ${\displaystyle \quad \quad {\frac {d^{k}}{dx^{k}}}x^{n}={\frac {n!}{(n-k)!}}\cdot x^{n-k}}$ Induction claim: ${\displaystyle \quad \quad \quad {\frac {d^{k+1}}{dx^{k+1}}}x^{n}={\frac {n!}{(n-k-1)!}}\cdot x^{n-k-1}}$ Induction step: ${\displaystyle \quad {\frac {d^{k+1}}{dx^{k+1}}}x^{n}=}$ The -th derivative is the derivative of the -th derivative: ${\displaystyle (k+1)}$ ${\displaystyle k}$ ${\displaystyle \quad {\frac {d}{dx}}{\frac {d^{k}}{dx^{k}}}x^{n}=}$ with the induction assumption: ${\displaystyle \quad {\frac {d}{dx}}{\frac {n!}{(n-k)!}}\cdot x^{n-k}=}$ ${\displaystyle \quad {\frac {n!}{(n-k)!}}\cdot (n-k)\cdot x^{n-k-1}=}$ ${\displaystyle \quad {\frac {n!}{(n-k-1)!}}\cdot x^{n-k-1}}$, q. e. d.

For some applications it is useful to define a function as the -th derivative of itself. As can be easily seen, the rule for also applies . ${\displaystyle 0}$ ${\displaystyle k=0}$

For is in particular ${\displaystyle k=n}$ ${\displaystyle {\frac {d^{n}}{dx^{n}}}x^{n}=n!}$

• ...For ${\displaystyle k>n:\quad {\frac {d^{k}}{dx^{k}}}x^{n}=0}$

This follows directly from the fact that the derivative of any constant function is the null function ; the latter also applies to the null function itself. ${\displaystyle {\frac {d^{n}}{dx^{n}}}x^{n}=n!}$

Individual evidence

1. ↑ Otto Forster : Analysis 1. Differential and integral calculus of a variable. Vieweg-Verlag (1976), ISBN 3-499-27024-2 , §15, example (15.9)
2. ↑ Otto Forster : Analysis 1. Differential and integral calculus of a variable. Vieweg-Verlag (1976), ISBN 3-499-27024-2 , §15, example (15.16)