13. Find dy/dx  in the following: 

y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right ) , -1 < x < 1

Answers (1)

Given function is
y = \cos ^{-1} \left ( \frac{2x}{1+ x^2 } \right )
We can rewrite it as
\cos y = \left ( \frac{2x}{1+ x^2 } \right )
Let's consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2x} )
Our equation reduces to 
\cos y = \sin 2t
Now, differentiation w.r.t. x is
\frac{d(\cos y)}{dx} = \frac{d(\sin2t)}{dt}.\frac{dt}{dx}
(-\sin y)\frac{dy}{dx} = 2(\cos 2t).\frac{1}{1+x^2} = \frac{2\cos2t}{1+x^2}= \frac{2.\frac{1-\tan^2 t}{1+\tan^2t}}{1+x^2} =\frac{2.\frac{1-x^2}{1+x^2}}{1+x^2} =\frac{2(1-x^2)}{(1+x^2)^2}
                                                                                                                                                               (\because \cos 2x = \frac{1-\tan^2 x}{1+\tan^2x} \ and \ x = \tan t)
\cos y = \ \left ( \frac{2 x }{1+ x^2 } \right )\Rightarrow \sin y = \frac{1-x^2}{1+x^2}
-\frac{1-x^2}{1+x^2}\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}
\frac{dy}{dx} = \frac{-2}{(1+x^2)}
Therefore, the answer is  \frac{-2}{1+x^2}

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