# 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

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

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.

 $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}y^{q}$ $\displaystyle=$ $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}x^{p}$ $\displaystyle\frac{\mathrm{d}}{\mathrm{d}y}(y^{q})\frac{\mathrm{d}y}{\mathrm{d% }x}$ $\displaystyle=$ $\displaystyle px^{p-1}$ $\displaystyle qy^{q-1}\frac{\mathrm{d}y}{\mathrm{d}x}$ $\displaystyle=$ $\displaystyle px^{p-1}$ $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}$ $\displaystyle=$ $\displaystyle\frac{p}{q}\frac{x^{p-1}}{y^{q-1}}$ $\displaystyle=$ $\displaystyle\frac{p}{q}x^{p-1}y^{1-q}$ $\displaystyle=$ $\displaystyle\frac{p}{q}x^{p-1}x^{p(1-q)/q}$ $\displaystyle=$ $\displaystyle\frac{p}{q}x^{p-1+p/q-p}$ $\displaystyle=$ $\displaystyle\frac{p}{q}x^{p/q-1}$

## 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

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

 $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(u^{n}u^{-n})$ $\displaystyle=$ $\displaystyle 1$ $\displaystyle u^{n}\frac{\mathrm{d}u^{-n}}{\mathrm{d}x}+u^{-n}\frac{\mathrm{d}% u^{n}}{\mathrm{d}x}$ $\displaystyle=$ $\displaystyle 0$ $\displaystyle u^{n}\frac{\mathrm{d}u^{-n}}{\mathrm{d}x}+u^{-n}(nu^{n-1})$ $\displaystyle=$ $\displaystyle 0$ $\displaystyle u^{n}\frac{\mathrm{d}u^{-n}}{\mathrm{d}x}$ $\displaystyle=$ $\displaystyle-nu^{-1}$ $\displaystyle\frac{\mathrm{d}u^{-n}}{\mathrm{d}x}$ $\displaystyle=$ $\displaystyle-nu^{-n-1}$
Title proof of the power rule ProofOfThePowerRule 2013-03-22 12:28:06 2013-03-22 12:28:06 mathcam (2727) mathcam (2727) 9 mathcam (2727) Proof msc 26A24 ProductRule Derivative2