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$$f(x) = \sqrt{1+\cot^2(x)}$$

How to calculate the derivative $f'(x)$? I've been looking at similar problems in my book and at examples, but I'm having a lot of trouble understanding it still. I'd appreciate it if someone could explain this in a way that would help not just answer this but other questions like it.

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We have $$ 1+\cot^2x=\frac{\sin^2x}{\sin^2x} + \frac{\cos^2x}{\sin^2x}\\ =\frac{1}{\sin^2x} $$ So $f(x) =1/\sqrt{\sin^2x} =1/|\sin x|$. Can you take it from there?

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  • $\begingroup$ You have to treat both cases, since sin can be either negative or positive. So on ]0, pi[ and ]pi, 2*pi[, sin being respectively positive and negative. $\endgroup$ – mvggz Oct 15 '14 at 22:36
  • $\begingroup$ The edit made my last comment useless, forget it $\endgroup$ – mvggz Oct 15 '14 at 22:37
  • $\begingroup$ Yes I know, but my computer does't put in screen, I've been having trouble with my connexion. I'll do it as soon as I get this fixed.. $\endgroup$ – mvggz Oct 15 '14 at 22:42
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Let $g(x) = \sqrt{x}, h(x) = 1 + x^2, k(x) = \cot x$. Then $f(x) = g(h(k(x)))$ and $\frac {df}{dx} = \frac{dg}{dh} \frac{dh}{dk} \frac{dk}{dx}$.
$$\frac{dg}{dh} = \frac{1}{2\sqrt{h(k(x))}}$$ $$\frac{dh}{dk} = 2k(x)$$ $$\frac{dk}{dx} = -\frac{1}{\sin^2x}$$ Can you continue by yourself from here?

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  • $\begingroup$ $-1/\sin^2x$ is the derivative of $\cot x$, not $1+\cot^2x$. $\endgroup$ – Barry Cipra Oct 15 '14 at 22:48
  • $\begingroup$ Oh, my apologies, will correct that now. $\endgroup$ – Andrei Rykhalski Oct 15 '14 at 22:49
  • $\begingroup$ OK, good. I would delete my comment, but for some reason the delete option for comments hasn't been showing up lately. The edit option and timestamps on comments have also disappeared. Has anyone else noticed this, or is it just me the site is messing with? $\endgroup$ – Barry Cipra Oct 15 '14 at 22:53
  • $\begingroup$ Not sure but perhaps it's impossible to delete comments with consequent comments after them. $\endgroup$ – Andrei Rykhalski Oct 15 '14 at 22:56
  • $\begingroup$ I don't think that's it. I've always been able to delete comments before, even after an interchange. Oddly enough, your last reply does have a timestamp, but the earlier ones don't (at least I'm not seeing them). And lo and behold, I am suddenly able to edit and/or delete this comment, which was also timestamped, but not my earlier comments! $\endgroup$ – Barry Cipra Oct 15 '14 at 23:00

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