# Inverse Trigonmetric functions

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• Dec 3rd 2007, 10:45 PM
Truthbetold
Inverse Trigonmetric functions
Find the dy/dx.

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

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

Thanks!
• Dec 3rd 2007, 10:58 PM
Jhevon
Quote:

Originally Posted by Truthbetold
Find the dy/dx.

arcsin ($\displaystyle \sqrt{x}$)
I got $\displaystyle \frac{\sqrt{x}}{\sqrt{1-2t}}$
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?

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

now, if $\displaystyle u = \sqrt{x}$, what is $\displaystyle u'$?
• Dec 3rd 2007, 11:12 PM
Truthbetold
Quote:

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

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

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

I don't know how I got t in my answer. I made a typo.

Derivative of u'= $\displaystyle \frac{1}{2\sqrt{x}}$
• Dec 3rd 2007, 11:14 PM
Jhevon
Quote:

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

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

ok, so now try again. what's the derivative of the original function?
• Dec 3rd 2007, 11:27 PM
Truthbetold
$\displaystyle \frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer
• Dec 3rd 2007, 11:28 PM
Jhevon
Quote:

Originally Posted by Truthbetold
$\displaystyle \frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer

that is correct. you could simplify it, but that's fine