Differentiation Problem.... (5)

Find the Derivative of $\tan^{-1} \left(\cfrac{\sqrt{1+x^2} -1}{x} \right) $ w.r.t $\cos^{-1} \left(\sqrt{\cfrac{1 + \sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right) $

Question

Differentiation Problem.... (5) 

Find the Derivative of $\tan^{-1} \left(\cfrac{\sqrt{1+x^2} -1}{x} \right) $ w.r.t $\cos^{-1} \left(\sqrt{\cfrac{1 + \sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right) $

mathslover · Answer

In order to give a brief idea about this... This question is from the topic - Differentiatin of Parametric Functions.

mathslover · Answer

And I'm not sure from where should I start dealing with it.

mathslover · Answer

@Callisto

anonymous · Answer

will
start like this 
$ (\tan^{-1} ( g(x) ))' = \tan'^{-1} g(x)   . g(x)'   $

anonymous · Answer

mmm chain rule

mathslover · Answer

Okay... tbh, I was not in the mood to try that. But, I will try this now with the help of ur suggestion... 1 minute

anonymous · Answer

ok shall i continue with u  ?
or u wanna do it by urself ?

mathslover · Answer

I will prefer that you continue with me... as I don't have proper plan in the mind to do it myself.

anonymous · Answer

ill give u the mood power :O
Goooooooooooooooo
|dw:1401801663735:dw|

anonymous · Answer

ok one thing i have to ask w.r.t cos....
means u wanna the answer with cos -1 ?

mathslover · Answer

lol...

mathslover · Answer

It is more like f(x) w.r.t g(x)

anonymous · Answer

hhh idk if that gonna work burt leet give it a try :)

anonymous · Answer

$\large (\tan^{-1} ( g(x) ))' = \tan'^{-1} g(x)   . g(x)' $
 =$\large  = \frac {1}{1+g(x)} . g'(x) $

anonymous · Answer

or u wanna proof that tan^-1 g(x)= 1/1+x ??

anonymous · Answer

idk why Os is laging for me -,,-

vishweshshrimali5 · Answer

Put, x = $\tan {\theta}$

vishweshshrimali5 · Answer

Then,

$$\tan^{-1} (\cfrac{\sqrt{1+x^2}-1}{x}) = \tan^{-1} (\cfrac{\sec{\theta} - 1}{\tan{\theta}})$$

vishweshshrimali5 · Answer

$$= \tan^{-1}(\cfrac{1-\cos{\theta}}{\sin{\theta}})$$

vishweshshrimali5 · Answer

Put, cos $\theta$ = 1-2$sin^2{\theta}$

mathslover · Answer

= theta/2 , right?

vishweshshrimali5 · Answer

*theta/2 in sine

vishweshshrimali5 · Answer

Yes

mathslover · Answer

Right, so, the next function is also theta/2 ?

mathslover · Answer

And thus : d(f(x))/d(g(x)) = 1 ? Right?

mathslover · Answer

f(x) = tan^{-1} ... 

g(x) = cos^{-1} ...

vishweshshrimali5 · Answer

Yes

mathslover · Answer

Got it.. Thanks a lot @vishweshshrimali5 . 

And thanks @BSwan .

anonymous · Answer

:D

mathslover · Answer

;)

anonymous · Answer

*head hurts* yup....time to go play a game to entertain my small attention span..