# Proof of d/dx (arccsc x)

1. Jan 11, 2009

### farleyknight

Hey all,

I'm trying to understand how $\frac{d}{dx} \mathrm{arccsc} x = -\frac{1}{|x| \sqrt{x^2 - 1}}$, which I found on Wikipedia, which is different than the form found in Stewart's calculus $\frac{d}{dx} \mathrm{arccsc} x = -\frac{1}{x \sqrt{x^2 - 1}}$. Where exactly does the |x| come from? There is proof of arcsin and arccos, but haven't yet found one which explains this detail for arccsc. Is there a proof that explains it?

Thanks,
- Farley

2. Jan 11, 2009

### jgens

I think it may be largely notational, because if we allow x < 0 than the derivative becomes indentical to d(arcsec(x))/dx.

Here's a proof for the derivative of arccsc(x):

csc(y) = x
d(csc(y))/dx = 1
-csc(y)cot(y)y' = 1
y' = -1/(csc(y)cot(y))
Now, since 1 + cot(x)^2 = csc(x)^2, cot^2(x) = csc^2(x) - 1, therefore:
y' = -1/(x(x^2-1)^1/2)

3. Jan 11, 2009

### lukaszh

$$\frac{\text{d}}{\text{d}x}\text{arccsc}x=\frac{\text{d}}{\text{d}x}\arcsin\left(\frac{1}{x}\right)$$
Substitution:
$$y=\arcsin u$$
Now I can use chain rule:
$$\frac{\text{d}y}{\text{d}u}\cdot\frac{\text{d}u}{\text{d}x}=\frac{1}{\sqrt{1-u^2}}\cdot\left(-\frac{1}{x^2}\right)=\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right)=\frac{1}{\sqrt{\frac{x^2-1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right)=\frac{|x|}{\sqrt{x^2-1}}\cdot\left(-\frac{1}{x^2}\right)$$
There is important to realise:
$$(x^2\geq0)\,\wedge\,(|x|\geq0)$$
so there quotient must be also positive.
$$\frac{|x|}{x^2}=\frac{1}{|x|}$$
If I use this to my proof, I obtain:
$$\frac{\text{d}}{\text{d}x}\text{arccsc}x=-\frac{1}{|x|\sqrt{x^2-1}}$$

Last edited: Jan 11, 2009
4. Jan 11, 2009

### jgens

Bah, an oversight on my part. I overlooked a restricted principle branch in the step following csc^2(x) - 1 = cot^2(x). Sorry!