logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Differentiate the following w.r.t x: $\tan^{-1}\Large {\bigg(\frac{\sqrt {1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\bigg)},\normalsize -1 < x < 1,x\neq 0$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $1+\cos x=2\cos^2\large\frac{x}{2}$
  • $1-\cos x=2\sin^2\large\frac{x}{2}$
  • $\large\frac{1-\tan x}{1+\tan x}$$=\tan\big(\large\frac{\pi}{4}$$-x)$
Step 1:
$y=\tan^{-1}\bigg[\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\bigg]$
Put $x^2=\cos\theta\Rightarrow \theta=\cos^{-1}x$
$\qquad\qquad=\tan^{-1}\bigg[\large\frac{\sqrt{1+\cos^2x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\bigg]$
But $1+\cos x=2\cos^2\large\frac{x}{2}$ and $1-\cos x=2\sin^2\large\frac{x}{2}$
$y=\tan^{-1}\bigg[\large\frac{\sqrt 2\cos \large\frac{x}{2}+\sqrt 2\sin\large\frac{x}{2}}{\sqrt 2\cos\Large\frac{x}{2}-\sqrt 2\sin\Large\frac{x}{2}}\bigg]$
$\;\;=\tan^{-1}\bigg[\large\frac{\cos x/2+\sin x/2}{\cos x/2-\sin x/2}\bigg]$
Step 2:
Divide each term by $\cos x/2$
$\;\;=\tan^{-1}\bigg[\large\frac{1+\tan\Large\frac{x}{2}}{1-\tan\Large\frac{x}{2}}\bigg]$
But $\tan\large\frac{\pi}{4}$$=1$
$\large\frac{1+\tan\large\frac{x}{2}}{1-\tan\large\frac{x}{2}}=\frac{\tan\Large\frac{\pi}{4}+\tan\Large\frac{x}{2}}{1-\tan\Large\frac{\pi}{4}\tan\Large\frac{x}{2}}$
This is of the form $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\qquad\;\;=\tan\big(\large\frac{\pi}{4}+\large\frac{x}{2})$
Therefore $y=\tan^{-1}\tan\big(\large\frac{\pi}{4}+\large\frac{x}{2})$
Step 3:
$y=\large\frac{\pi}{4}+\frac{x}{2}$
Now differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{1}{2}$
answered Jul 1, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...