Thread: 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..

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

Originally Posted by Jameson

Yes. So y' = arctan'[(arcsin(2x))] * arcsin'[(2x)] * 2

Thank you so much.

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..

So now use the chain rule like you alluded to.

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

To the both of you, thank you so much.