Differentiability of $x^{\frac{1}{n}}$

Question

I want to show that the function $x^\frac{1}{n}$ with $n\in\mathbb{N}^*$ is differentiable for all $x>0$

Let $x_{0}\in\mathbb{R}^{+*}$

$\cfrac{f(x)-f(x_{0})}{x-x_{0}}=\cfrac{x^{\frac{1}{n}}-{x_{0}}^{\frac{1}{n}}}{x-x_{0}}$

I don't know where to go from there, I need some kind of algebraic manipulation in order to take the limit. Any tips?

Answer 1

Hint: For $x,x_0$ strictly positive, we have $$ x-x_0=(x^{1/n})^n-(x_0^{1/n})^n=\cdots $$

Answer 2

Hint. One may use that, for $n=2,3,\cdots$, $$ a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right). $$

Answer 3

Since $$x - x_0 = (x^{1/n})^n-(x_0^{1/n})^n = (x^{1/n} - x_0^{1/n})((x^{1/n})^{n-1} + (x^{1/n})^{n-2} (x_0^{1/n})^1 + \dots + (x_0^{1/n})^{n-1}),$$ $$\frac{f(x)-f(x_{0})}{x-x_{0}}=\frac{x^{1/n}-{x_{0}}^{1/n}}{x-x_{0}}=\frac{1}{(x^{1/n})^{n-1} + (x^{1/n})^{n-2} (x_0^{1/n})^1 + \dots + (x_0^{1/n})^{n-1}},$$ so $$\lim_{x \to x_0} \frac{f(x)-f(x_{0})}{x-x_{0}}=\frac{1}{(x^{1/n})^{n-1} + (x^{1/n})^{n-2} (x_0^{1/n})^1 + \dots + (x_0^{1/n})^{n-1}} = \frac{1}{n x_0^{(n-1)/n}}.$$