# derivative of ${x}^{n}$

Recall the typical derivative^{} formula^{} is

$$\frac{df}{dx}=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}.$$ |

The derivative of ${x}^{n}$ can be computed directly from this formula utilizing the binomial theorem, see for instance an alternative proof of the deriviative of ${x}^{n}$ (http://planetmath.org/AlternativeProofOfDerivativeOfXn).

However, to avoid invoking the binomial theorem one can often make use of alternative definitions of the derivative which are justified by inspecting a diagrams and/or through the use of algebra. Instead of using $h$, use $h=x-a$ so that $a\to x$ is the same as $h\to 0$. This gives the formula

$$\frac{df}{dx}=\underset{a\to x}{lim}\frac{f(x)-f(a)}{x-a}.$$ |

This is the standard slope formula between the two points $(a,f(a))$ and $(x,f(x))$ only now we let $a$ approach $x$.

From this formula the casual rule

$$\frac{d}{dx}({x}^{n})=n{x}^{n-1}$$ |

for positive integer values of $n$ can be easily derived.

First notice that

$$(x-a)({x}^{n-1}+{x}^{n-2}a+\mathrm{\cdots}+x{a}^{n-2}+{a}^{n-1})={x}^{n}-{a}^{n}.$$ |

Therefore

$$\frac{d}{dx}({x}^{n})=\underset{a\to x}{lim}\frac{{x}^{n}-{a}^{n}}{x-a}=\underset{a\to x}{lim}({x}^{n-1}+{x}^{n-2}a+\mathrm{\cdots}+x{a}^{n-2}+{a}^{n-1})=n{x}^{n-1}.$$ |

When $n$ is not a positive integer the proof typically depends on implicit differentiation^{} as follows:

$$y={x}^{n};\mathrm{ln}y=\mathrm{ln}{x}^{n}=n\mathrm{ln}x;\frac{{y}^{\prime}}{y}=n\frac{1}{x};{y}^{\prime}=n\frac{y}{x}=n{x}^{n-1}.$$ |

For the theoretically inclined this solution can be disappointing because it depends on a proof for the derivative of $\mathrm{ln}x$. Most texts simply redefine $\mathrm{ln}x$ as the integral^{} of $1/x$ or in some similar fashion delay an honest proof.

For this reason it is often instructive to prove the power rule^{} in stages depending on the type of exponents. Having proven the result for $n$ a positive integer, one can extend this to $-n$ using the product rule^{}.

To begin with observe $1={x}^{n}{x}^{-n}$. Therefore

$0$ | $=$ | $\frac{d}{dx}}(1)={\displaystyle \frac{d}{dx}}({x}^{n}{x}^{-n})={\displaystyle \frac{d}{dx}}({x}^{n}){x}^{-n}+{x}^{n}{\displaystyle \frac{d}{dx}}({x}^{-n})$ | ||

$=$ | $n{x}^{n-1}{x}^{-n}+{x}^{n}{\displaystyle \frac{d}{dx}}({x}^{-n})={\displaystyle \frac{n}{x}}+{x}^{n}{\displaystyle \frac{d}{dx}}({x}^{-n}).$ |

Now solve for $\frac{d}{dx}({x}^{-n})$.

$$\frac{d}{dx}({x}^{-n})=-\frac{n}{x}\frac{1}{{x}^{n}}=(-n){x}^{(-n)-1}.$$ |

Likewise fractional powers can also be accommodated without the use of $\mathrm{ln}x$ by beginning with the property: given $1/b$ a rational number^{} then

$$x={({x}^{1/b})}^{b}.$$ |

Using the chain rule^{} we can prove the power rule for ${x}^{1/b}$ as follows.

$$1=\frac{d}{dx}(x)=\frac{d}{dx}({({x}^{1/b})}^{b})=b{({x}^{1/b})}^{b-1}\frac{d}{dx}({x}^{1/b}).$$ |

Once again solve for $\frac{d}{dx}({x}^{1/b})$

$$\frac{d}{dx}({x}^{1/b})=\frac{1}{b{({x}^{1/b})}^{b-1}}=\frac{1}{b}{x}^{1/b}{x}^{-1}=\frac{1}{b}{x}^{1/b-1}.$$ |

Finally, the derivative of ${x}^{a/b}$ for any fraction $a/b$ is done by observing that ${x}^{a/b}={({x}^{a})}^{1/b}$ so indeed the chain rule once again solves the problem.

Title | derivative of ${x}^{n}$ |
---|---|

Canonical name | DerivativeOfXn |

Date of creation | 2013-03-22 15:50:10 |

Last modified on | 2013-03-22 15:50:10 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 16 |

Author | Algeboy (12884) |

Entry type | Theorem |

Classification | msc 26B05 |

Classification | msc 26A24 |

Synonym | Power rule |

Related topic | DerivativesByPureAlgebra |

Related topic | AlternativeProofOfDerivativeOfXn |