proof of the power rule

The power rule can be derived by repeated application of the product rule.

Proof for all positive integers $n$

The power rule has been shown to hold for $n=0$ and $n=1$.If the power rule is known to hold for some $k>0$, then we have

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}{x}^{k+1}$$=$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}(x\cdot x^k)$

$=$$x\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}{x}^k\right)+x^k$

$=$$x\cdot (k{x}^{k-1})+x^k$

$=$$k{x}^k+x^k$

$=$$(k+1)x^k$

Thus the power rule holds for all positive integers $n$.

Proof for all positive rationals $n$

Let $y=x^{p/q}$. We need to show

$$\frac{\mathrm{d}y}{\mathrm{d}x}(x^{p/q})=\frac{p}{q}{x}^{p/q-1}$$(1)

The proof of this comes from implicit differentiation.

By definition, we have $y^q=x^p$. We now take the derivative with respect to $x$ on both sides of the equality.

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}{y}^q$$=$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}{x}^p$

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}y}}(y^q){\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$$=$$p{x}^{p-1}$

$q{y}^{q-1}{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$$=$$p{x}^{p-1}$

${\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$$=$${\displaystyle \frac{p}{q}}{\displaystyle \frac{x^{p-1}}{y^{q-1}}}$

$=$${\displaystyle \frac{p}{q}}{x}^{p-1}{y}^{1-q}$

$=$${\displaystyle \frac{p}{q}}{x}^{p-1}{x}^{p(1-q)/q}$

$=$${\displaystyle \frac{p}{q}}{x}^{p-1+p/q-p}$

$=$${\displaystyle \frac{p}{q}}{x}^{p/q-1}$

Proof for all positive irrationals $n$

For positive irrationals we claim continuity due to the fact that (1)holds for all positive rationals, and there are positive rationals that approach any positive irrational.

Proof for negative powers $n$

We again employ implicit differentiation. Let $u=x$, and differentiate $u^n$ with respect to $x$ for some non-negative $n$. We must show

$$\frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}=-n{u}^{-n-1}$$(2)

By definition we have $u^n{u}^{-n}=1$.We begin by taking the derivative with respect to $x$ on both sides of the equality. By application of the product rule we get

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}(u^n{u}^{-n})$$=$$1$

$u^n{\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}+u^{-n}{\displaystyle \frac{\mathrm{d}{u}^n}{\mathrm{d}x}}$$=$$0$

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

$u^n{\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}$$=$$-n{u}^{-1}$

${\displaystyle \frac{\mathrm{d}{u}^{-n}}{\mathrm{d}x}}$$=$$-n{u}^{-n-1}$