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  • Find dy/dx  in the following: y = tan ^ -1 3x - x ^3/ 1 - 3x ^2

10. Find dy/dx  in the following: 
y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right ) , - \frac{1}{\sqrt 3 } < x < \frac{1}{\sqrt 3 }

Answers (1)

best_answer

Given function is
y = \tan ^{-1} \left ( \frac{3x- x ^3}{1- 3x ^2} \right )
Lets 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{3x-x^3}{1-3x^2} = \frac{3\tan t-\tan^3t }{1-3\tan^2t} = \tan3t \ \ \ \ \ \ (\because \tan 3x = \frac{3\tan x-\tan^3x }{1-3\tan^2x} )
Our equation reduces to 
y = \tan^{-1}(\tan 3t)
y = 3t
Now, differentiation w.r.t. x is
\frac{d(y)}{dx} = \frac{d(3t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 3.\frac{1}{1+x^2} = \frac{3}{1+x^2}
Therefore, the answer is  \frac{3}{1+x^2}

Posted by

Gautam harsolia

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