Question

find the derivative of
tan inverse ( (sqrt(1+x^2)+sqrt(1-x^2) ) / ( (sqrt(1+x^2)-sqrt(1-x^2) )

Answer 1

multiply by the conjugate

Answer 2

i am sure there is some tremendous simplification involved before taking the derivative.
try and put x= tan y
\(\huge \dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\dfrac{1+\sqrt{\dfrac{1-x^2}{1+x^2}}}{1-\sqrt{\dfrac{1-x^2}{1+x^2}}}\)

with x= tan y
\(\sqrt{\dfrac{1-x^2}{1+x^2}}=\sqrt{\dfrac{1-\tan^2y}{1+\tan^2y}}=...?\)
can you simplify this ?

Answer 3

no i'm not getting this simplified but long route did give me answer as - x / sqrt(1-x^4)

Answer 4

you tried with conjugate right ?

Answer 5

actually, x^2 =sin y will be a better substitution

Answer 6

\( \large
\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \frac{1 - \tan(\theta/2)}{1 + \tan(\theta/2)} \)

Answer 7

no i tried directly diff for tan inverse with chain rule

Answer 8

with x^2 = sin y
\(\huge \dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\dfrac{1+\sqrt{\dfrac{1-x^2}{1+x^2}}}{1-\sqrt{\dfrac{1-x^2}{1+x^2}}} \\ \huge \dfrac{1+\dfrac{1-\tan(y/2)}{1+\tan(y/2)}}{1-\dfrac{1-\tan(y/2)}{1+\tan(y/2)}}=\dfrac{1}{\tan(y/2)} \\ \huge \tan^{-1}(\cot(y/2))=\pi/2-y/2 = \\\huge =\pi/2-\sin^{-1}(x^2)/2\)
see, how simplified it became :)

Answer 9

and then one step derivative of sin inverse will give you required answer, same as what u got :)

Answer 10

got how i simplified?
ask if any doubts :)

Answer 11

u said x^2 = sin y, then how did u get tan y/2

Answer 12

1-siny = sin^y/2+cos^y/2-2siny/2*cosy/2 =(cosy/2-siny/2) ^2
similarly 1+siny = (cosy/2+siny/2)^2

Answer 13

thanks a lot

Answer 14

you are welcome..