Question

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

Answer 1

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))`

Answer 2

`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))`