Proof of the Power Rule

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The Power Rule
If $a$ is any real number, and $f(x) = x^a,$ then

The proof is divided into several steps. However, you can skip to the last step for a quick proof that uses the formula for the derivative of exponential functions.
Step 1: Proof of the Power Rule for NonNegative Integer Exponents
In this step, we assume that $f(x) = x^n,$ where $n$ is some positive integer: $0, 1, 2, 3,$ .... First, if n happens to be zero, $f(x) = x^0 = 1,$ a constant, and so its derivative is zero, by the result we proved in the text. Thus, assume that $n$ is a positive integer. To follow the proof of this step you must recognize a nice little algebraic fact. First look at these identities.(Use the distributive law to expand the righthand side in each case.)
These examples generalize to give us the following formula
Difference of Two n th Powers
If $a$ and $b$ are real numbers, and $n$ is a positive integer, then

$f^{'}(x)$  $=$  $\lim_{h \to 0}\frac{f(x+h)  f(x)}{h}$ 
$=$  $\lim_{h \to 0}\frac{(x+h)^n  x^n}{h}$ 
$=$  $\lim_{h \to 0}\frac{[(x+h)  x][(x+h)^{n1} + (x+h)^{n2} x + . . . + x^{n1}]}{h}$  
$=$  $\lim_{h \to 0}\frac{h[(x+h)^{n1} + (x+h)^{n2} x + . . . + x^{n1}]}{h}$  Since $(x+h)  x = h$  
$=$  $\lim_{h \to 0}[(x+h)^{n1} + (x+h)^{n2} x + . . . + x^{n1}]$  Cancel the $h$ 
Step 2: Proof of the Power Rule for Negative Integer Exponents
Here, we assume that $f(x) = x6{n},$ where $n = 0, 1, 2, 3,....$ Since we are required here to justify the power rule for negative integers, we can't simply go ahead and use it! "Officially," all we can use are the power rule for positive integers (Step 1), and the product and quotient rules. (The product rule is proved in the text, and the quotient rule is proved here.) What we can do is this: write $x^{n}$ as $\frac{1}{x^{n}}$ and use the quotient rule. Applying the quotient rule to $\frac{1}{x^{n}}$ gives$=$  $\frac{(0)(x^n)  (1)(nx^{n1})}{(x^n)^2}$  
$=$  $\frac{nx^{n1}}{x^{2n}}$  
$=$  $ nx^{n12n} = nx^{n1},$ 
Step 3: Proof of the Power Rule for Rational Exponents
For this step, we assume that $f(x) = x^{p/q},$ where $p$ and $q$ are integers (positive or negative), and we must show that
whereas
Equating these derivatives gives
$\frac{d}{dx} (x^{\frac{p}{q}})$  $=$  $\frac{px^{p1}}{q(x^{\frac{p}{q}})^{q1}}$ 
$=$  $\frac{px^{p1}}{qx^{p \frac{p}{q}}}$  
$=$  $\frac{p}{q} x^{p1 (p \frac{p}{q})}$  
$=$  $\frac{p}{q} x^{\frac{p}{q}1}.$ 
Step 4: Proof of the Power Rule for Arbitrary Real Exponents (The General Case)
Actually, this step does not even require the previous steps, although it does rely on the use of exponential functions and their derivatives.
First, we need the equality $x^n = e^{nlnx}.$ (You can check this by taking the natural logarithm of both sides).
$=$  $\frac{d}{dx} e^{nlnx}$  
$=$  $e^{nlnx} \frac{d}{dx} [n lnx]$  By the derivative rule for exponential functions  
$=$  $e^{nlnx} \left\[ \frac {n}{x} \right\]$  By the derivative rule for logarithmic functions  
$=$  $x^{n} \left\[ \frac{n}{x} \right\]$  By the equality $x^n = e^{nlnx}$  
$=$  $nx^{n1}.$ 
This proves the power rule for all real powers.