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)

Re: Another derivatives of inverse trig function

$\displaystyle y = arcsec(2x^2)$

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

$\displaystyle sec(y)=2x^2$

Re: Another derivatives of inverse trig function

Quote:

Originally Posted by

**Quacky** $\displaystyle y = arcsec(2x^2)$

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

$\displaystyle sec(y)=2x^2$

with that I get

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

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

From here I do not know where to go.

Re: Another derivatives of inverse trig function

Quote:

Originally Posted by

**Quacky** $\displaystyle y = arcsec(2x^2)$

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

$\displaystyle sec(y)=2x^2$

Quote:

Originally Posted by

**Barthayn** with that I get

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

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

From here I do not know where to go.

Let's see.

$\displaystyle sec(y)=2x^2$

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

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

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

Using the identity:

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

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

We had:

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

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

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

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

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 :)

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)