Christmas Streak 33/88: What About Real Numbers?

Calculus Level 3

The formula $\dfrac{d}{dx}\left(x^k\right)=kx^{k-1}~(x\ge 0)$ can be proved easily via binomial expansion if $k$ is a natural number.

Does the formula also hold if $k$ is a real number but not an integer?

Yes, and the derivative is always defined over all reals No, it never holds Yes, but for some

k,

the derivative is not defined for some

x.

The humanity would never figure out the answer No, not always even though it holds for some

k

1 solution

Boi (보이)
Nov 1, 2017

Answer: Yes, but for some $k,$ the derivative is not defined for some $x.$

The formula can be proved by using the derivative of the natural log, and the chain rule.

Let $f(x)=x^k.$

$\ln |f(x)|=k\ln|x| \\ \frac{f'(x)}{f(x)}=\frac{k}{x} \\ f'(x)=\frac{k}{x}\cdot f(x) = kx^{k-1}.$

The formula is proven, however, this might not be defined when $x=0.$

One example is $f(x)=x^{\frac{1}{2}}=\sqrt{x}.$ The derivative is $f'(x)=\dfrac{1}{2\sqrt{x}},$ however this isn't defined for $x=0.$