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Question: Calculate the derivatives of the following functions (specify domain):

a. $\lfloor x^2 \rfloor \sin^2(\pi x)$

b. {$x^2 $}$\sin^2(\pi x)$

What I did:

on a. If $x \in \Bbb N$ then the function equals 0 (because of the sine). If $x \notin \Bbb N$ then I need to calculate the limit $\lim_{h \to 0}\frac {f(x+h)-f(x)}h$ which I'm having trouble calculating, as both the numerator and denominator are going to be 0.

on b. I divided to the same cases and when x is an integer it's also 0. But this time I think I can just say that it's differentiable for $x \in \Bbb R \setminus \Bbb N$ and simply remove the braces in my calculation. Am I correct?

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  • $\begingroup$ Is that greatest integer function? $\endgroup$ – Neer Jun 5 '13 at 6:35
  • $\begingroup$ floor is greatest integer below x. {} is the same as x-floor(x) $\endgroup$ – jreing Jun 5 '13 at 6:51
  • $\begingroup$ For $ n < x < n+1 $ both of floor and fractional part of $x^2$ have discontinuities. $\endgroup$ – M. Strochyk Jun 5 '13 at 6:58
  • $\begingroup$ could you please elaborate $\endgroup$ – jreing Jun 5 '13 at 7:08
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    $\begingroup$ For example, if $ 2\leqslant x<3 $ then $4 \leqslant x^2<9,$ so $\{\sqrt{5},\ \sqrt{6},\ \sqrt{7},\ \sqrt{8}\}\subset(2,\ 3)$ and $$\lfloor x^2 \rfloor=\begin{cases}4,& 2\leqslant x<\sqrt{5}, \\ 5,& \sqrt{5}\leqslant x<\sqrt{6},\\ 6,& \sqrt{6}\leqslant x<\sqrt{7},\\ 7,& \sqrt{7}\leqslant x<\sqrt{8}, \\ 8,& \sqrt{8}\leqslant x<3.\end{cases}$$ $\endgroup$ – M. Strochyk Jun 5 '13 at 7:25
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First, some comments on your work so far:

  • For part (a), your observation that the function is $0$ when $x \in \mathbb{N}$ is true, but irrelevant. You don't care about the function's value -- you only care about its derivative, and whether or not it exists.

  • For part (b), you are incorrect (though I think you may mean something different than what you said). Again you cannot rule out the case $x \in \mathbb{N}$ simply by observing that the function takes on the value $0$. Moreover, the function is not differentiable in all of $\mathbb{R} \setminus \mathbb{N}$.

Now, here are some hints.

  • For part (a), note that if $a^2 \in \mathbb{N}$, then $f(x) = \lfloor x^2 \rfloor \sin^2 (\pi x)$ is not continuous at $x = a$ (why?); therefore, $f$ is not differentiable at $x = a$. Then, in the case where $a^2 \not \in \mathbb{N}$, observe that $\lfloor x^2 \rfloor$ is constant on a some small interval containing $a$. Therefore, you can just treat $\lfloor x^2 \rfloor$ as a constant when computing $f'(a)$.

  • For part (b), you can take an easy shortcut. Just write

$$ \{x^2\}\sin^2(\pi x) = x^2 \sin^2(\pi x) - \lfloor x^2 \rfloor \sin^2(\pi x) $$

You know the derivative of $x^2 \sin^2(\pi x)$ and that it exists everywhere. Also, you know the derivative of $\lfloor x^2 \rfloor \sin^2(\pi x)$ and exactly when it exists. So you should be done.

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  • $\begingroup$ for part (a) what your saying about calculating $f'(a)$ - should I do it with the limit definition? I don't know what is the limit of the expression I get there... $\endgroup$ – jreing Jun 5 '13 at 13:59
  • $\begingroup$ @user1685224 So, you shouldn't need the limit definition. Since $\lfloor x^2 \rfloor$ is constant in the small interval you're worried about, you only need to know the derivative of $\sin^2(\pi x)$. Do you know how to compute the derivative of $\sin^2 (\pi x)$? $\endgroup$ – 6005 Jun 5 '13 at 14:17
  • $\begingroup$ Without the limit definition yes, But I thought I need to prove it using the limit definition. Or maybe I'm missing out on what is the justification that allowes me to use the "regular" derivative of sin. $\endgroup$ – jreing Jun 5 '13 at 14:26
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    $\begingroup$ You don't have to go all the way back to the limit definition every time you want the derivative to exist. This would be a pain, which is why tools like the product rule and chain rule were developed. The chain rule theorem says that if $f(x)$ is differentiable and $g(x)$ is differentiable, then $f(g(x))$ is differentiable too. In particular, let $f(x) = x^2$ and $g(x) = \sin(\pi x)$. $\endgroup$ – 6005 Jun 5 '13 at 14:36
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    $\begingroup$ @user1685224 yes, that is correct. $\endgroup$ – 6005 Jun 5 '13 at 14:41

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