I'm having a hard time with the following problem: y=arctan(arcsin(2x)) Am I supposed to do chain rule? Someone please help..
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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.. $\displaystyle \frac{d}{dx}\bigg[\arctan(x)\bigg]=\frac{1}{1+x^2}$ $\displaystyle \frac{d}{dx}\bigg[\arcsin(2x)\bigg]=\frac{2}{\sqrt{1-4x^2}}$ So now use the chain rule like you alluded to.
Last edited by Mathstud28; Nov 3rd 2008 at 06:19 PM. Reason: typo
Yesssss, I used the actual derivatives when I solved the problem, not just arctan'. To the both of you, thank you so much.
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