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

Aug 6, 2014

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$

Process:

1.) $y = \text{arccsc} \left(x\right)$

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.) $\frac{1}{\sin} y = x$

Solve for $y$:

4.) $1 = x \sin y$

5.) $\frac{1}{x} = \sin y$

6.) $y = \arcsin \left(\frac{1}{x}\right)$

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

We know that $\frac{d}{\mathrm{dx}} \left[\arcsin \alpha\right] = \frac{1}{\sqrt{1 - {\alpha}^{2}}}$ (there is a proof of this identity located here)

So, take the derivative of the outside function, then multiply by the derivative of $\frac{1}{x}$:

7.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]$

The derivative of $\frac{1}{x}$ is the same as the derivative of ${x}^{- 1}$:

8.) $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot \left(- {x}^{- 2}\right)$

Simplifying 8. gives us:

9.) $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{{x}^{2} \cdot \sqrt{1 - \frac{1}{x} ^ 2}}$

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

10.) $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{{x}^{4} \left(1 - \frac{1}{x} ^ 2\right)}}$

Simplifying yields:

11.) $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$

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.