# derivative of $x^{n}$

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

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 $a\rightarrow x$ is the same as $h\rightarrow 0$. This gives the formula

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

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

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

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

First notice that

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

Therefore

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

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

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

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 $1/x$ or in some similar fashion delay an honest proof.

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

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

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

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

Likewise fractional powers can also be accommodated without the use of $\ln x$ by beginning with the property: given $1/b$ 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 $x^{n}$ DerivativeOfXn 2013-03-22 15:50:10 2013-03-22 15:50:10 Algeboy (12884) Algeboy (12884) 16 Algeboy (12884) Theorem msc 26B05 msc 26A24 Power rule DerivativesByPureAlgebra AlternativeProofOfDerivativeOfXn