Got an interesting problem

If $y = \tan^{-1} \left( \cfrac{\sqrt{1+x} - \sqrt{1-x} }{\sqrt{1+x} + \sqrt{1-x} } \right) $ , then $y'(0) = ?$

Question

Got an interesting problem

If $y = \tan^{-1} \left( \cfrac{\sqrt{1+x} - \sqrt{1-x} }{\sqrt{1+x} + \sqrt{1-x} } \right) $ , then $y'(0) = ?$

hartnn · Answer

i sense a great simplification with a substitution

goformit100 · Answer

it's a 2014 year CBSE board exam question.

hartnn · Answer

x = cos 2t

mathslover · Answer

@hartnn - 

I can solve it further with your suggestion of substitution of x = cos 2t 

But,I want to ask that how to predict whether this substitution will work or not?

mathslover · Answer

Like sometimes, people suggest to substitute x = tan theta

I tried to substitute that here, but it led me to again an indeterminate form. Moreover, I tried rationalizing it too.. but that didn't work too.

hartnn · Answer

1+ something and 1- something both are present

in entire trigonometry
there is only cos 2x formula which will simplify that

anonymous · Answer

i think hart nailed it ;-)

mathslover · Answer

Oh, so that was how you predicted the substitution...! Great. Yeah @mukushla ... I'm really angry with Hartnn currently.. I thought, no one would be able to solve it :(

goformit100 · Answer

divide numerator and denominator by 
under root ( 1 + x)

mathslover · Answer

But still,  @hartnn but thanks for that suggestion. I will keep in mind how to predict the substitution . This would be really helpfull for me in future.

mathslover · Answer

I will close it now.. and find the toughest question for @hartnn ... :P Just joking.

goformit100 · Answer

Refer to NCERT and See solution from http://www.meritnation.com/