What is the derivative of `f(x)=csc^-1(x)` ?

`dy/dx = -1/sqrt(x^4 - x^2)`

Process:

1.) `y = "arccsc"(x)`

First we will rewrite the equation in a form that is easier to work with.

Take the cosecant of both sides:

2.) `csc y = x`

Rewrite in terms of sine:

3.) `1/siny = x`

Solve for `y`:

4.) `1 = xsin y`

5.) `1/x = sin y`

6.) `y = arcsin (1/x)`

Now, taking the derivative should be easier. It's now just a matter of chain rule.

We know that `d/dx[arcsin alpha] = 1/sqrt(1 - alpha^2)` (there is a proof of this identity located here)

So, take the derivative of the outside function, then multiply by the derivative of `1/x`:

7.) `dy/dx = 1/sqrt(1 - (1/x)^2) * d/dx[1/x]`

The derivative of `1/x` is the same as the derivative of `x^(-1)`:

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

Simplifying 8. gives us:

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

To make the statement a little prettier, we can bring the square of `x^2` inside the radical, although this isn't necessary:

10.) `dy/dx = -1/(sqrt(x^4(1 - 1/x^2)))`

Simplifying yields:

11.) `dy/dx = -1/sqrt(x^4 - x^2)`

And there is our answer. Remember, derivatives problems involving inverse trig functions are mostly an exercise in your knowledge of trig identities. Use them to break down the function into a form that's easy to differentiate.