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

1 Answer
Aug 6, 2014

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.