Derivatives of inverse trig functions

[mhunt1739]

Find dy/dx at the indicated point for the equation-

arcsin xy = 2/3 arctan 2x,

The point is (1/2,1)

I am not sure how to go about doing this. I know the equation for d/dx of arcsin u = (u'/ sqrt 1-u^2) and arctan u is u'/1+u^2 and i am stumped from there on.

 

[Deleted member 4993]

Find dy/dx at the indicated point for the equation-

arcsin xy = 2/3 arctan 2x,

The point is (1/2,1)

I am not sure how to go about doing this. I know the equation for d/dx of arcsin u = (u'/ sqrt 1-u^2) and arctan u is u'/1+u^2 and i am stumped from there on.

Iff the problem has been posted correctly - then apply chain rule and march on:

\(\displaystyle \frac{d(\sin^{-1}xy)}{dx} = \frac{2}{3}\frac {d(\tan^{-1}2x)}{dx}\)

\(\displaystyle \frac{1}{\sqrt{1-(xy)^2}}\cdot [y + xy'] = \frac{2}{3}\cdot\frac {1}{1 + 4x^2}\cdot 2\)

Now isolate y'....