What is the derivative of #y=arccsc(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.