Question

How do you find the derivative of `tan(arcsin(x))`?

Answer 1

`d/(dx) tan(arcsin(x))= 1/(1-x^2)^(3/2)`

Explanation:

Let `t = arcsin(x)`

Then:

`x = sin(t)`

So:

`tan(arcsin(x)) = tan(t) = sin(t)/cos(t) = x/sqrt(1-x^2)`

So:

`d/(dx) tan(arcsin(x))`

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

`= (1-x^2)^(-1/2) + x*(-1/2)(1-x^2)^(-3/2)*(-2x)`

`= (1-x^2)(1-x^2)^(-3/2) +x^2(1-x^2)^(-3/2)`

`= 1/(1-x^2)^(3/2)`