# Math Help - Trouble with the derivative of inverse trigonometric functions.

1. ## Trouble with the derivative of inverse trigonometric functions.

I'm having a hard time with the following problem:

y=arctan(arcsin(2x))

Am I supposed to do chain rule?
Someone please help..

2. Originally Posted by MadHotThunder
I'm having a hard time with the following problem:

y=arctan(arcsin(2x))

Am I supposed to do chain rule?
Someone please help..
Yes. So y' = arctan'[(arcsin(2x))] * arcsin'[(2x)] * 2

3. Originally Posted by Jameson
Yes. So y' = arctan'[(arcsin(2x))] * arcsin'[(2x)] * 2
Thank you so much.

4. Originally Posted by MadHotThunder
I'm having a hard time with the following problem:

y=arctan(arcsin(2x))

Am I supposed to do chain rule?
Someone please help..
$\frac{d}{dx}\bigg[\arctan(x)\bigg]=\frac{1}{1+x^2}$

$\frac{d}{dx}\bigg[\arcsin(2x)\bigg]=\frac{2}{\sqrt{1-4x^2}}$

So now use the chain rule like you alluded to.

5. Yesssss, I used the actual derivatives when I solved the problem, not just arctan'.

To the both of you, thank you so much.