# How do you find the derivative of #y=arcsin(1/x)#?

##### 2 Answers

First, recall the identity

If this identity doesn't look familiar then I may recommend viewing a few videos from this page as they present a couple identities like this, and explain why they are true.

Differentiating

The derivative of

Now, all we need to do is simplify a bit:

#### Explanation:

We may also know from the outset that:

#d/dx"arcsec"(x)=1/(absxsqrt(x^2-1))# #d/dx"arccsc"(x)=-1/(absxsqrt(x^2-1))#

Then:

#y=arcsin(1/x)#

#sin(y)=1/x#

#1/(sin(y))=x#

#csc(y)=x#

#y="arccsc"(x)#

Then:

#dy/dx=-1/(absxsqrt(x^2-1))#