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

2 Answers
Jul 31, 2014

First, recall the identity #d/dx[arcsinalpha] = 1/(sqrt(1 - x^2))#.

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 #arcsin (1/x)# is just a matter of using the identity above, as well as the chain rule:

#dy/dx = 1/sqrt(1-(1/x)^2) * d/dx[1/x]#

The derivative of #1/x# is found using the power rule:

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

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

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

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

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

Oct 2, 2016

Answer:

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

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))#