Proof of $f'(x) = \frac{1}{n} \cdot x^{\left(\frac{1}{n}\right) - 1}$ when $f(x) = x^{\frac{1}{n}}$ by limit definition

Question

how can I prove

$$f'(x) = \frac{1}{n} \cdot x^{\left(\frac{1}{n}\right) - 1}$$ when $$f(x) = x^{\frac{1}{n}}$$ by limit definition?

Thank you.

hint: $x^u-x_0^u=(x-x_0)(x^{u-1}+x^{u-2}x_0+x^{u-3}x_0^2+ \cdots + xx_0^{u-2}+x_0^{u-1})$ — randomgirl, Apr 21 '15 at 21:18
Closely related: Differentiation using first principles with rational powers — Dave L. Renfro, Apr 21 '15 at 21:54

score 2 · Answer 1 · edited Apr 21 '15 at 21:26

2

$$\lim_{x\to x_0}\frac{x^{1/n}-x_0^{1/n}}{x-x_0}=\lim_{x\to x_0}\frac{x-x_0}{(x-x_0)\left(\left(x^{1/n}\right)^{n-1}+\left(x^{1/n}\right)^{n-2}x_0+\ldots+\left(x_0^{1/n}\right)^{n-1}\right)}=$$

$$=\frac1{\left(x_0^{1/n}\right)^{n-1}+\left(x_0^{1/n}\right)^{n-2}x_0+\ldots+\left(x_0^{1/n}\right)^{n-1}}=\frac1{n\,x_0^{1-1/n}}=\frac{x_0^{1/n-1}}n$$

edited Apr 21 '15 at 21:26

Mark Viola

179,405

answered Apr 21 '15 at 21:20

Timbuc

34,191

Is there a way to prove through $h \to 0$ ? – bru1987 Apr 21 '15 at 21:24
@bru1987 Just let $x = x_0+h$ then $x\to x_0\iff h\to 0$. – Eff Apr 21 '15 at 21:39

Proof of $f'(x) = \frac{1}{n} \cdot x^{\left(\frac{1}{n}\right) - 1}$ when $f(x) = x^{\frac{1}{n}}$ by limit definition

1 Answers1