1

The problem in my textbook asks me to find the derivative of the following.

  1. $y = \tan^{-1}\frac{3x - x^3}{1 - 3x^2}, \frac{-1}{\sqrt3} < x < \frac{1}{\sqrt3}$

I get here that the restraints are present for $x$ because if $x$ were equal to $\frac{-1}{\sqrt3}$ or $\frac{1}{\sqrt3}$ then the denominator would be $0$ and $y$ will be undefined.

  1. $y = \cos^{-1}\frac{1 - x^2}{1 + x^2}, 0 < x < 1$

Here, I don't understand why $0$ and $1$ aren't included. When $x = 1$, $y = \cos^{-1}(0)$ and when $x = 0$, $y = \cos^{-1}(1)$ both of which are defined right? So could someone please explain why the domain doesn't include $1$ and $0$ in the second question?

Alain Remillard
  • 3,625
user662650
  • 525
  • Does the textbook ever ask for a derivative of a function defined on a closed (or half-closed) interval? It may have to do with the way the book defined the derivative. – David K Apr 07 '20 at 12:27
  • No. All the questions have domains like this where the endpoints aren't included. – user662650 Apr 07 '20 at 12:56
  • What book are you using? The clue is probable somewhere else in the textbook. Your second function is well define for all $x$. – Alain Remillard Apr 07 '20 at 13:03
  • It's the NCERT 12th grade math book. This is the standard book used across the nation. – user662650 Apr 07 '20 at 13:05
  • If "derivative" is defined such that you need both a left and right derivative, you can only have a derivative at a number in the interior of the domain; if the number is the maximum of the domain then you can't have a right derivative there. It is possible to define a valid derivative anyway, but I would guess the author did not believe it was necessary to do so. – David K Apr 07 '20 at 13:22

2 Answers2

1

Has someone pointed out un a comment to another answer, function does not have to be define on its largest possible domain.

In your examples, the domain is larger than what is define. First function $$y=\tan^{-1}\frac{3x-x^3}{1-3x^2}$$ is define everywhere except for $x=\pm\frac1{\sqrt3}$, since the arctangente is define everywhere.

Second function is define everywhere since $$-1<\frac{1-x^2}{1+x^2}\leq1$$ for all $x\in\Bbb R$.

Alain Remillard
  • 3,625
0

If you're asking whether there's any specific reason for the inequalities to be strict then in the case of the upper bound (1) i can't see one (but then why should there be?). As for the lower bound (0), $cos^{-1}$ isn't differentiable at 1, and so $y$ isn't differentiable at x=0.

Mr_Random_Guy
  • 390
  • 1
    You should also understand that a function does not HAVE TO be defined on its largest possible domain. For example, I can, if I wish, define f(x) to be "$f(x)= x^2$ for 0< x< 1" only. – user247327 Apr 07 '20 at 12:24
  • When x = 1, y = 0. $f'(x) = \frac{2}{1+x^2}$ and $f'(x)$ at x =1, is 1. So it is differentiable at x = 1 right? – user662650 Apr 07 '20 at 12:29