I have tried to solve it but I don't why it's wrong. I need to take the derivative of $\;\operatorname{arccoth}(\sin x)$:

By using chain rule, I get: $$\dfrac 1{1 - \sin^2 x}\cdot \cos x = \dfrac {\cos x}{\cos^2 x} = \dfrac{1}{\cos x} = \sec x.$$

But the answer in the book is $-\csc x$.

Why ?


You're correct that by the chain rule, we have $$\frac d{dx}\Big(\operatorname{arccoth}(\sin x)\Big) = \dfrac 1{1 - \sin^2 x}\cdot \cos x = \dfrac {\cos x}{\cos^2 x} = \dfrac{1}{\cos x} = \sec x.$$

The answer in the book may be mis-numbered, intending to refer to another problem of the same sort. Note that if the problem were to find the derivative of $\,\operatorname {arccoth}( \cos x)$, then you can confirm that the answer would be $-\csc x$.

But the correct answer to your posted problem of finding the derivative of $\;\operatorname{arccoth}(\sin x)\,$ is, indeed, $\sec x.$

  • $\begingroup$ Could use another UV +1 $\endgroup$ – Amzoti Dec 5 '13 at 2:02

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