# Thread: Find the second derivative of

1. ## Find the second derivative of

$y=\frac{\sqrt{1-x^2}}{x}$

This is what I get so far:
$y'=\frac{-x-\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}$

y''= I'm working on it but I'm kinda confused.

Is y' correct or did I mess up?

2. Originally Posted by yoman360
$y=\frac{\sqrt{1-x^2}}{x}$

This is what I get so far:
$y'=\frac{-x-\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}$

y''= I'm working on it but I'm kinda confused.

Is y' correct or did I mess up?
I hate the quotient rule and will do anything I can to avoid it.

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

$\frac{d}{dx}\left[\sqrt{x^{-2}-1}\right]=\frac{1}{2\sqrt{x^{-2}-1}}\cdot\frac{-2}{x^3}=\frac{-1}{x^3\sqrt{x^{-2}-1}}=\frac{-1}{\sqrt{x^4-x^6}}=-(x^4-x^6)^{-1/2}$

$\frac{d}{dx}\left[-(x^4-x^6)^{-1/2}\right]=\frac{1}{2}(x^4-x^6)^{-3/2}\cdot(4x^3-6x^5)=\boxed{\frac{2x^3-3x^5}{(x^4-x^6)^{3/2}}}$

3. Actually, you can simplify it even more.

$\frac{2x^3-3x^5}{(x^4-x^6)^{3/2}}=\frac{2x^3-3x^5}{[x^2(x^2-x^4)]^{3/2}}=\frac{2x^3-3x^5}{x^3(x^2-x^4)^{3/2}}=\boxed{\frac{2-3x^2}{(x^2-x^4)^{3/2}}}$