Tracker Posted January 28, 2009 Share Posted January 28, 2009 I am trying to prove the follow two questions without using the differentiation formulas for inverse trig. or inverse hyperbolic functions. [math] \arcsin(u) = \arctan(\frac{u}{\sqrt{1 - u^2}}) [/math] I have been looking at it as [math] \frac{opp}{adj} = \frac{opp}{hyp} [/math] but something in my steps is off. I am also trying to solve [math] \frac{d}{dx}[\tan(\arcsin(x))] = \frac{1}{(1 - x^2)^\frac{3}{2}} [/math] Link to comment Share on other sites More sharing options...
timo Posted January 28, 2009 Share Posted January 28, 2009 For the first one consider a triagle with side-lengths 1, u, [math] \sqrt{1-u^2}[/math]. For the 2nd one use the result of the first one. Be sure to check the ranges, btw. Sidenote: I did not try them myself but these two approaches seem to be the obvious ones. Link to comment Share on other sites More sharing options...
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