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

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))