0
$\begingroup$

Let $f(x)=\left\{\begin{array}{c}x^2\sin(1/x), x\ne 0\\0, x=0\end{array}\right.$
Why is it possible to calculate the derivative of the function when $x\neq 0$ "regularly" using the chain rule, and Lebiniz rule, but for the case when $x=0$ you have to calculate it by the definition of the derivative instead, and not just say that the derivative of a constant function is $0$?

$\endgroup$
4
  • 1
    $\begingroup$ The function would have to be constant on an open interval containing $0$ in order for that argument to be valid. $\endgroup$ – David Mitra Jan 18 at 13:17
  • $\begingroup$ Perhaps I am missing something here; I am confused by your query. Using the standard definition of the derivative, the chain rule is routinely proven. Thus, (in effect), in all cases, the derivative is based on the standard definition $\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}.$ $\endgroup$ – user2661923 Jan 18 at 13:19
  • $\begingroup$ I see by the other responses that I have (somewhat) interpreted your query backwards. That is, instead of answering why you can't regard the function as constant at $x=0$, I have answered that the chain rule is nothing more than the standard definition in disguise. Therefore, in all cases, the derivative is based on the standard definition. $\endgroup$ – user2661923 Jan 18 at 13:23
  • 1
    $\begingroup$ When you are given the definition of a function using cases (ie different formulas for different range of values of $x$) then it does not necessarily mean that derivative can be obtained by differentiating those formulas for each case. One actually needs to clearly see what cases are given. Let's not be too mechanical and instead apply definitions and theorems related to derivatives. $\endgroup$ – Paramanand Singh Jan 18 at 14:02
2
$\begingroup$

Because this is NOT a "constant function"! The fact that $f(0)= 0$ does not make it a "constant function". Every function has some specific value at $x= 0$ (or at any $x$). The fact that the value happens to be $0$ does not matter.

$\endgroup$
1
$\begingroup$

Note that the limit of your function is defined as \begin{equation} f^{\prime}\left(0\right) = \lim_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x} = \lim_{x\to 0}\frac{x^{2}\sin{1/x}}{x} = \lim_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x} = \lim_{x\to 0}{x\sin{1/x}}. \end{equation} Besides, a constant typically means a constant over a neighborhood. It is meaningless to say "constant over a single point". Your function only says that it is $0$ at $0$. It is not a "constant" in any neighborhood around $0$.

$\endgroup$
1
  • $\begingroup$ +1 there. I want to add that value of a function at a point is always a fixed number and if we go by the logic (in question) then derivative should be zero everywhere. The notion of derivative (and all things in calculus in general) always needs values at the point under consideration as well as values near that point. $\endgroup$ – Paramanand Singh Jan 18 at 14:05
0
$\begingroup$

"...and not just say that the derivative of a constant function is $0$".

There is no interval $(-\delta,\delta)$ in which the function is equal to the constant $0$ for all $x$ in this interval.

$\endgroup$
0
$\begingroup$

Let $g(x)=\left\{\begin{array}{c}99x+x^2\sin(1/x), x\ne 0\\0, x=0\end{array}\right.$

By your logic, the derivative of $g$ at $0$ is $0$. But I hope you can see that the derivative is actually $99$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.