Math Help - Inverse Trigonmetric functions

1. Inverse Trigonmetric functions

Find the dy/dx.

arcsin ( $\sqrt{x}$)
I got $\frac{\sqrt{x}}{\sqrt{1-2t}}$
This, is not correct,

I used the typical trig formula for this. 1/(1-u^2) * du/dx.

Thanks!

2. Originally Posted by Truthbetold
Find the dy/dx.

arcsin ( $\sqrt{x}$)
I got $\frac{\sqrt{x}}{\sqrt{1-2t}}$
This, is not correct,

I used the typical trig formula for this. 1/(1-u^2) * du/dx.

Thanks!
now how in the world did you end up with t in your answer?

$\frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}}$ where $u$ is a function of $x$

now, if $u = \sqrt{x}$, what is $u'$?

3. Originally Posted by Jhevon
now how in the world did you end up with t in your answer?

$\frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}}$ where $u$ is a function of $x$

now, if $u = \sqrt{x}$, what is $u'$?
I don't know how I got t in my answer. I made a typo.

Derivative of u'= $\frac{1}{2\sqrt{x}}$

4. Originally Posted by Truthbetold
I don't know how I got t in my answer. I made a typo.

Derivative of u'= $\frac{1}{2\sqrt{x}}$
ok, so now try again. what's the derivative of the original function?

5. $\frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer

6. Originally Posted by Truthbetold
$\frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer
that is correct. you could simplify it, but that's fine