Question

what is the Differentiation of (sec-1x)^2 ?

Answer 1

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

Explanation:

We want to find the derivative of

`y=(sec^-1(x))^2`

Use the Chain Rule if `y=f(g(x))=f(u)`

then `dy/dx=dy/(du)*(du)/dx`

Let `y=u^2` so that `u=sec^-1(x)`

Then `dy/(du)=2u` and `(du)/dx=1/(x^2sqrt(1-x^-2))`*

*Optionally see explanation

Apply the Chain Rule

`dy/dx=2u1/(x^2sqrt(1-x^-2))`

Substitute `u=sec^-1(x)`

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