# Another derivatives of inverse trig function

• November 11th 2011, 04:20 PM
Barthayn
Another derivatives of inverse trig function
I need help taking the derivative of y = arcsec(2x^2)
This is what I do
y = arcsec(2x^2)
x=sec(2x^2)
y'=-4ysin(2y^2)

I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help? This seems not to make any sense at all for me (Headbang)
• November 11th 2011, 04:22 PM
Quacky
Re: Another derivatives of inverse trig function
$y = arcsec(2x^2)$

$sec(y)=sec(arcsec(2x^2))$

$sec(y)=2x^2$
• November 11th 2011, 04:30 PM
Barthayn
Re: Another derivatives of inverse trig function
Quote:

Originally Posted by Quacky
$y = arcsec(2x^2)$

$sec(y)=sec(arcsec(2x^2))$

$sec(y)=2x^2$

with that I get

$y'= \frac{4x}{2x^2tany}$

$y'= \frac{2 cot(y)}{x}$

From here I do not know where to go.
• November 11th 2011, 04:46 PM
Quacky
Re: Another derivatives of inverse trig function
Quote:

Originally Posted by Quacky
$y = arcsec(2x^2)$

$sec(y)=sec(arcsec(2x^2))$

$sec(y)=2x^2$

Quote:

Originally Posted by Barthayn
with that I get

$y'= \frac{4x}{2x^2tany}$

$y'= \frac{2 cot(y)}{x}$

From here I do not know where to go.

Let's see.

$sec(y)=2x^2$

$sec(y)tan(y)y'=4x$

$y'=\frac{4x}{sec(y)tan(y)}$

$y'= \frac{4x}{2x^2tan(y)}$ as you had (sorry, I had to get there for myself)

Using the identity:

$tan^2(y)+1=sec^2(y)$, we get that:

$tan(y)=\sqrt{sec^2(y)-1}$

$y'= \frac{4x}{2x^2tan(y)}$

$y'= \frac{4x}{2x^2\sqrt{sec^2(y)-1}}$

$y'= \frac{4x}{2x^2\sqrt{4x^4-1}}$

$y'= \frac{2}{x\sqrt{4x^4-1}}$
• November 11th 2011, 04:49 PM
Barthayn
Re: Another derivatives of inverse trig function
Thanks, I had you third last step and see sec^2y and I was all like "what do I do? Help someone!". Thanks for pointing out the unit trig identity. It was much easier than it appeared :)
• November 11th 2011, 05:02 PM
Quacky
Re: Another derivatives of inverse trig function
Quote:

Originally Posted by Barthayn
Thanks, I had you third last step and see sec^2y and I was all like "what do I do? Help someone!". Thanks for pointing out the unit trig identity. It was much easier than it appeared :)

Isn't this always the case?(Wink)