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How can I calculate the derivative in $x=0$ of $$f(x)=(x-\sin x)^{1/3}?$$ I found the derivative and replaced $x$ but it would be an indeterminate form and if I try using limit of $(f(x)-f(0))/x$, it doesn't lead me anywhere so how could I find it?

Mee Seong Im
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Lola
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4 Answers4

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The formula for the derivative doesn't make sense at $0$. Indeed, for $x\ne0$, $$ f'(x)=\frac{1}{3}(x-\sin x)^{-2/3}(1-\cos x) $$ but the expression is undefined for $x=0$. The reason is that $x\mapsto x^{1/3}$ is not differentiable at $0$; however, this doesn't mean the function you're given isn't differentiable at $0$.

You have two ways out of the dilemma. One is to compute $$ f'(0)=\lim_{x\to0}f'(x) $$ (this is possible by l'Hôpital), the other is to use the definition: $$ f'(0)=\lim_{x\to0}\frac{(x-\sin x)^{1/3}}{x}= \lim_{x\to0}\left(\frac{x-\sin x}{x^3}\right)^{\!1/3} $$ Can you find $$ \lim_{x\to0}\frac{x-\sin x}{x^3} $$ with Taylor expansion or l'Hôpital?

egreg
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$$\left(\frac{f(x)-f(0)}x\right)^3=\frac{x-\sin x}{x^3}.$$

By L'Hospital, twice,

$$\frac{1-\cos x}{3x^2}$$ then $$\frac{\sin x}{6x}.$$

Can you conclude ?

Without L'Hospital, write

$$L=\lim_{x\to0}\frac{(x-\sin x)^{1/3}}x=\lim_{x\to0}\frac{(3x-\sin 3x)^{1/3}}{3x}=\lim_{x\to0}\left(\frac{3x-3\sin x}{27x^3}+\frac{4\sin^3x}{27x^3}\right)^{1/3}=\left(\frac{L^3}9+\frac4{27}\right)^{1/3}.$$

You can draw $L$.

  • yes the answer is $\frac{1}{6^{1/3}} $ ,didnt know I am allowed to raise the limit at 3 and then apply the square root – Lola Jul 10 '17 at 16:20
  • @Lola: yep, the limit of a constant power is the constant power of the limit, because the limit of a product is the product of the limits. I used the trick to ease applicaiton of L'Hospital. –  Jul 10 '17 at 16:22
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HINT: the first derivative is given by $$f'(x)=1/3\,{\frac {1-\cos \left( x \right) }{ \left( x-\sin \left( x \right) \right) ^{2/3}}} $$ and the limit for $x$ tends to zero exists and is equal to $$\frac{1}{6^{1/3}}$$

Dr. Sonnhard Graubner
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Use Taylor: $$f'(0)=\lim_{x\to 0}\frac{(x-\sin x)^{1/3}-0}x.$$ In which \begin{align} \sin x &=x-\frac16x^3+O(x^6)\\ (x-\sin x)^{1/3}&=(\frac16x^3+O(x^6))^{1/3}=(\frac16)^{1/3}x(1+O(x^3))^{1/3} \end{align} Can you take it from here?

Vim
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