Derivative of inverse cosecant? I am slightly confused by this, because when I worked out the derivative of arccosec(x), my answer was $\frac{-1}{x\sqrt{x^2-1}}$, which agrees with the answers online. However this would imply that the gradient of the arccosec fubction is positive for negative values of x, however looking at sketches of arccosecx it seems to have negative gradient for both positive and negative x. Does this has something to do with taking the positive square root of x^2-1 rather than the negative root? Thank you in advance.

Your intuition is almost correct, but not quite. You can think about this in two ways, depending on how you define $\operatorname{arccosec}$

Firstly, note that $\operatorname{arccosec}{x}= \arcsin{\left(\frac{1}{x}\right)}$ (this is one possible definition). Then taking the derivative gives $$ \frac{d}{dx} \operatorname{arccosec}{x} = \frac{-1}{x^2}\arcsin'{\left(\frac{1}{x}\right)} = -\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}, $$ and we can see that this is negative for either sign of $x$ ($|x|\geqslant1$, of course). Taking out the factor of $1/\sqrt{x^2}$ from the denominator requires choosing a sign for the square root.

On the other hand, you can do this using the inverse function rule as $$ \operatorname{arccosec}'{x} = \frac{1}{\operatorname{cosec}'{\operatorname{arccosec}{x}}} = \frac{1}{x\cot{\operatorname{arccosec}{x}}}, $$ and again, you have $\operatorname{cosec}^2{x} = 1+\cot^2{x}$, but have to choose the sign correctly.