Derivative of a function at a point

Question

A friend who is still in highschool gave me the following problem. For $x>-1\,$ compute $f'(0)$ if $$f(x)=\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}$$ Here is how I solved it. Since it just got messier if I computed the derivative then tried to plug in$x=0$ I used the definition of derivate: $$\lim_{x\to 0}\frac{\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}-\left(0-0\right)^\frac{1}{3}}{x-0}=\lim_{x\to 0}\frac{\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}}{x}.$$ We have the power series of $\ln(1+x)=x-\frac{x^2}{2}+O(x^3)\rightarrow\ln^2(1+x)=x^2-x^3+O(x^4)$ so the initial limit is just $$\lim_{x\to 0}\frac{\left(x^3(1-O(x)\right)^\frac{1}{3}}{x}=1$$ I couldn't solve it with any other method and my friend didnt learn series yet. Could you help me with an elementary approach for this problem?

Answer 1

$$\lim_{x\to 0}\frac{\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}}{x}=l$$ Then by algebraic limit theorem $$\lim_{x\to 0}\frac{\left(x^2-\ln^2(x+1)\right)}{x^3}=l^3$$ $$\lim_{x\to 0}\frac{\left(x-\ln(x+1)\right)}{x^2}\lim_{x\to 0}\frac{\left(x+\ln(x+1)\right)}{x}=l^3$$ Now using L'Hopital's rule on both parts $$\lim_{x\to 0}\frac{\left(1-\frac1{(x+1)}\right)}{2x}\cdot\lim_{x\to 0}\frac{\left(1+\frac{1}{x+1}\right)}{1}=l^3 \\ \lim_{x\to0}\frac{x}{2x(x+1)}\cdot\lim_{x\to0}\frac{x+2}{x+1}=l^3 \\ l^3=1 \\l=1$$

Answer 2

This is a rearranged version of Piyush Divyanakar's answer without the assumption that the given limit exists.

Note that by L'Hopital, $$\lim_{x\to 0}\frac{x+\ln(x+1)}{x}=\lim_{x\to 0}\frac{1+1/(x+1)}{1}=2$$ and $$\lim_{x\to 0}\frac{x-\ln(x+1)}{x^2}=\lim_{x\to 0}\frac{1-1/(x+1)}{2x}=\lim_{x\to 0}\frac{1/(x+1)^2}{2}=\frac{1}{2}.$$ Hence, as $x\to 0$, $$\frac{\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}}{x}=\left(\frac{x+\ln(x+1)}{x}\cdot \frac{x-\ln(x+1)}{x^2}\right)^{1/3}\to \left(2\cdot \frac{1}{2}\right)^{1/3}=1.$$ Hence $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\frac{\left(x^2-\ln^2(x+1)\right)^\frac{1}{3}}{x}=1.$$