I have to differentiate $\sin^{-1}(\sqrt{\csc{x}})$ with respect to $x$. Used the inverse trig formula/identity to get $(\sqrt{\csc{x}})^2=\csc{x}.$

So then $\frac{1}{\sqrt{1-\csc{x}}}$. This I multiplied by the derivative of the square root of $\csc{x}$ getting $$\frac{1}{\sqrt{1-\csc{x}}}*-\frac{1}{2}*\csc{x}^{1/2}*\cot{x}$$

The answer given for the problem is saying $$\frac{1}{\sqrt{1-\csc{x}}}*-\frac{1}{2}*\csc(x)*\cot x$$ though.

  • 2
    $\begingroup$ Common operators and standard functions often have their own $\TeX$-command, such as \sin and \csc. This is also explained in this basic MatJaX tutorial. $\endgroup$
    – gebruiker
    Dec 14, 2015 at 18:33
  • 1
    $\begingroup$ For me, your work is very correct. We can bet that, once more, typo's in a textbook ? $\endgroup$ Dec 14, 2015 at 18:39
  • 1
    $\begingroup$ You are right and the book is wrong. $\endgroup$
    – rogerl
    Dec 14, 2015 at 18:40
  • $\begingroup$ Alright, thanks guys. I seem to run into that a fair bit in this course. It's not a book, it's all stuff the teacher has made up. Some time's I'm not quite sure if the mistake is with that or I'm just doing something stupid and not noticing when there's small differences between the answer and what I get like with this. $\endgroup$
    – windy401
    Dec 14, 2015 at 18:46
  • $\begingroup$ Since you asked the question, you cannot be stupid. So, how many hypotheses are left ? Cheers :-) $\endgroup$ Dec 14, 2015 at 18:53

1 Answer 1



Using the chain rule, $\frac{\text{d}}{\text{d}x}\left(\arcsin\left(\sqrt{\csc(x)}\right)\right)=\frac{\text{d}\arcsin(u)}{\text{d}u}\cdot\frac{\text{d}u}{\text{d}x}$

where $u=\sqrt{\csc(x)}$ and $\frac{\text{d}}{\text{d}u}\left(\arcsin(u)\right)=\frac{1}{\sqrt{1-u^2}}$:


Using the chain rule, $\frac{\text{d}}{\text{d}x}\left(\sqrt{\csc(x)}\right)=\frac{\text{d}\sqrt{u}}{\text{d}u}\cdot\frac{\text{d}u}{\text{d}x}$

Where $u=\csc(x)$ and $\frac{\text{d}}{\text{d}u}\left(\sqrt{u}\right)=\frac{1}{2\sqrt{u}}$:


  • $\begingroup$ You may want to look at the final result there again. $\endgroup$
    – windy401
    Dec 14, 2015 at 19:01

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.