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I have this function: $f(x)=(\cos(x)-1)/(x^{1/3}+1)$ which is continuous at $0$ and I wanted to know if it's differentiable at $0$.

The way I tried to solved it was using the definition:

$$\lim_{h\to 0} \frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{\cos(h)-1}{h}\cdot\frac{1}{h^{1/3}+1}=0$$

I found that for $0+$ and $0-$ the limit is $0$. Doesn't this mean that it's differentiable at $0$?

But when I calculate the derivative:

$$f'(x)=-\frac{\sin(x)}{x^{1/3}+1}-\frac{\cos(x)-1}{3\cdot x^{2/3}\cdot(x^{1/3}+1)^2}$$

I found that $f'(x)$ is not continuous at $0$. Doesn't this mean that it's not differentiable at $0$?

C Squared
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Christihan
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    There are differentiable functions whose derivatives are not continuous. Example: $F(x)=\int_{-1}^x\text{sgn}(t)dt$. Upon differentiating, $F'(x)$ is not continuous at $0$. – C Squared May 07 '21 at 07:23
  • you might find this extract useful. https://community.middlebury.edu/~abbott/UA/UA-5-1.pdf – J.Zelez May 07 '21 at 07:51
  • $f$ is infinitely differentiable at $x=0$. –  May 07 '21 at 08:06
  • @CSquared: the question is not about continuous differentiability, it is about differentiability. –  May 07 '21 at 09:54
  • @YvesDaoust OP said that when they calculated the derivative of $f$, they found that $f'$ was discontinuous at $0$, and then they thought that meant $f$ was not differentiable at $0$. I was trying to clarify that functions can have a discontinuous first derivative while still being differentiable. – C Squared May 07 '21 at 10:45

2 Answers2

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Continuity in the formal derivation of $f'(x)$ is not of significant importance, but having limit is. The derivative of this function has a limit equal to zero at $x=0$, so the problem of discontinuity of $f(x)$ in $x=0$ can be removed by manually defining $f'(0)=\lim_{x\to 0}f'(x)$.

Mostafa Ayaz
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  • This answer is misleading. The function does have the derivative $0$ at $x=0$, you don't need to remove a discontinuity. –  May 07 '21 at 07:36
  • Then how do you cope with $f'(0)$ from the second term of the derivative formula, knowing that it is equal to $\frac{0}{0}$? – Mostafa Ayaz May 07 '21 at 07:38
  • Please see my answer. By the way, it is completely obvious that the given function has a smooth maximum at $x=0$, and is even infinitely differentiable there. –  May 07 '21 at 07:41
  • @YvesDaoust, I think this edit must do the job... – Mostafa Ayaz May 07 '21 at 07:44
  • Stay hooked to your point of view if you want, but the function is differentiable, full stop. There is no concept of "analytical continuity". –  May 07 '21 at 07:46
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Indeed,

$$\lim_{h\to0}\dfrac{\dfrac{\cos h-1}{h^{1/3}+1}-0}h=0$$ and $f'(0)=0$.

Having an indeterminate in the analytical expression is not a reason for the derivative not to exist, it just shows that the analytical approach is not bulletproof.

  • what do you mean by not bullet proof? – C Squared May 07 '21 at 10:50
  • im not sure what point you're trying to make and was asking if you could clarify what you meant by 'the analytical approach' not being bulletproof. I wouldn't have asked if it was obvious, as i see no holes in this method. the limit can still be computed analytically. – C Squared May 07 '21 at 11:41
  • thank you for not answering my question and assuming that I don't understand derivatives. it was a simple question. – C Squared May 07 '21 at 12:01