Question

What is the derivative of `arcsin(1/x)`?

Answer 1

`(dy)/dx=(-1)/(xsqrt(x^2-1)`

Explanation:

To find `d/dx(arcsin(1/x))`
`arcsin(1/x)=sin^-1(1/x)`
Let `y=sin^=1(1/x)`
Let `u=1/x`
`(du)/dx=(-1)/x^2`
`y=sin^-1u`
`(dy)/dx=1/sqrt(1-u^2)(du)/dx`
Substituting the values of `u and (du)/dx`
we have

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

Simplifying

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

`(dy)/dx=(-1)/(xsqrt(x^2-1)`