PlanetMath: derivative of $x^n$

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derivative of $x^n$

(Theorem)

Recall the typical derivative formula is

$\displaystyle \frac{df}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.$

The derivative of

can be computed directly from this formula utilizing the binomial theorem, see for instance an alternative proof of the deriviative of

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 , use so that $a\rightarrow x$ is the same as $h\rightarrow 0$ . This gives the formula

$\displaystyle \frac{df}{dx}=\lim_{a\rightarrow x}\frac{f(x)-f(a)}{x-a}.$

This is the standard slope formula between the two points

and

only now we let

approach

From this formula the casual rule

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

for positive integer values of

can be easily derived.

First notice that

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

Therefore

$\displaystyle \frac{d}{dx}(x^n)=\lim_{a\rightarrow x} \frac{x^n-a^n}{x-a} =\lim_{a\rightarrow x} (x^{n-1}+x^{n-2}a+\cdots + xa^{n-2}+ a^{n-1}) =nx^{n-1}.$

When is not a positive integer the proof typically depends on implicit differentiation as follows:

$\displaystyle y=x^n;\quad \ln y=\ln x^n=n\ln x;\quad \frac{y'}{y}=n\frac{1}{x};\quad y'=n\frac{y}{x}=nx^{n-1}.$

For the theoretically inclined this solution can be disappointing because it depends on a proof for the derivative of $\ln x$ . Most texts simply redefine $\ln x$ as the integral of

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 a positive integer, one can extend this to using the product rule.

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

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

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

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

Likewise fractional powers can also be accommodated without the use of $\ln x$ by beginning with the property: given a rational number then

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

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

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

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

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

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

"derivative of $x^n$

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Power rule

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Cross-references: fraction, chain rule, rational number, property, fractional powers, product rule, exponents, type, similar, integral, solution, implicit differentiation, integer, positive, points, slope, algebra, definitions, binomial theorem, derivative
There are 6 references to this entry.

This is version 13 of derivative of $x^n$

, born on 2006-04-08, modified 2006-09-06.
Object id is 7811, canonical name is DerivativeOfXn.
Accessed 3300 times total.

Classification:

AMS MSC:	26A24 (Real functions :: Functions of one variable :: Differentiation : general theory, generalized derivatives, mean-value theorems)
	26B05 (Real functions :: Functions of several variables :: Continuity and differentiation questions)

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