Question

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

Answer 1

`-1/(xsqrt(x^2-1))`

Explanation:

To differentiate this we will be applying a chain rule:

Start by Letting `theta=arcsin(1/x)`

`=>sin(theta)=1/x`

Now differentiate each term on both sides of the equation with respect to `x`

`=>cos(theta)*(d(theta))/(dx)=-1/x^2`

Using the identity : `cos^2theta+sin^2theta=1 => costheta=sqrt(1-sin^2theta)`

`=>sqrt(1-sin^2theta)*(d(theta))/(dx)=-1/x^2`

`=>(d(theta))/(dx)=-1/x^2*1/sqrt(1-sin^2theta)`

Recall : `sin(theta)=1/x" "` and `" "theta=arcsin(1/x)`

So we can write,

`(d(arcsin(1/x)))/(dx)=-1/x^2*1/sqrt(1-(1/x)^2)=-1/x^2*1/sqrt((x^2-1)/x^2)`

`=-1/x^2*x/sqrt(x^2-1)=color(blue)(-1/(xsqrt(x^2-1)))" or "-sqrt(x^2-1)/(x(x^2-1))`