1. ## derivative

I am trying to take the derivative of this:
y = arccsc^2(x)
I know this can be simplied as arccsc(x)^2
I am trying to find y(prime)..can I use natural logs to do that?
Could I do lny = 2ln(arccsc(x))
Thanks for the feedback.

2. You could perhaps try it this way:

$\displaystyle \frac{d}{dx}[csc^{-1}(x)]$

$\displaystyle y=csc^{-1}(x), \;\ x=csc(y)$

Differentiate:

$\displaystyle \frac{d}{dx}{x}=\frac{d}{dx}[csc(y)]\Rightarrow{1=-csc(y)cot(y)}\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{1}{-csc(y)cot(y)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{-csc(csc^{-1}(x))cot(csc^{-1}(x))}$

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

$\displaystyle \frac{d}{dx}[csc^{-1}(x)^{2}]$

Chain rule:

$\displaystyle 2csc^{-1}(x)\frac{d}{dx}[csc^{-1}(x)]$

$\displaystyle \boxed{\frac{-2csc^{-1}(x)}{x\sqrt{x^{2}-1}}}$

3. Originally Posted by galactus
You could perhaps try it this way:

$\displaystyle \frac{d}{dx}[csc^{-1}(x)]$

$\displaystyle y=csc^{-1}(x), \;\ x=csc(y)$

Differentiate:

$\displaystyle \frac{d}{dx}{x}=\frac{d}{dx}[csc(y)]\Rightarrow{1=-csc(y)cot(y)}\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{1}{-csc(y)cot(y)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{-csc(csc^{-1}(x))cot(csc^{-1}(x))}$

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

$\displaystyle \frac{d}{dx}[csc^{-1}(x)^{2}]$

Chain rule:

$\displaystyle 2csc^{-1}(x)\frac{d}{dx}[csc^{-1}(x)]$

$\displaystyle \boxed{\frac{-2csc^{-1}(x)}{x\sqrt{x^{2}-1}}}$
I'm not really understanding how do did this. You found the derivative of arccsc(x), (1/-x((squareroot(x^2 - 1)) But how did you figure out yprime from y = arccsc^2(x)?

4. The last part. The chain rule.

5. Ohh..ok
Where i am getting lost is how are you breaking y = arccsc^2 (x) up?
I see you take the derivative of arccsc(x) (which you set to y)
but where is the (csc^-1 (x)^2) coming from?