PlanetMath: proof of the power rule

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proof of the power rule

(Proof)

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

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

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

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

(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.

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

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

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

(2)

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

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