Statement
The derivative of $ x^r $ is $ r * x^{r-1} $, where $ r $ is a real number.
$$ \tfrac{d}{dx}(x^r) = r * x^{r-1}, r \in \Reals $$
Proof
Define function $ f $.
$$ f(x) = x^r $$
Rewrite the function using $ e $ and $ \ln $. Since they are inverse functions, it stays exactly the same.
$$ f(x) = e^{\ln(x^r)} $$
Differentiate this function by the derivative of e^x, and mutliply by the chain rule.
$$ f'(x) = e^{\ln(x^r)} * \tfrac{d}{dx}(\ln(x^r)) $$
The first factor simplifies to $ x^r $, since $ e $ and $ \ln $ are inverses.
$$ f'(x) = x^r * \tfrac{d}{dx}(\ln(x^r)) $$
For the second factor, bring down the exponent $ r $ using the properties of logarithms. Then bring $ r $ out of the $ \frac{d}{dx} $ function.
$$ f'(x) = x^r * r * \tfrac{d}{dx}(\ln(x)) $$
The derivative of $ \ln(x) $ is $ \frac{1}{x} $.
$$ f'(x) = x^r * r * \frac{1}{x} $$
Now simplify this function to the final result by using negative exponents and adding exponents.
$$ f'(x) = r * x^r * x^{-1} $$
$$ f'(x) = r * x^{r - 1} $$