Question 1

Find dy/dx of the following

1) y = tan^-1(3x-x³/1-3x²) where -1/√3 < x < 1/√3

Question 2

${ \boxed{ \purple{ \sf{ (dy)/(dx) = \frac{3}{1 + {x}^(2)}}}}}$

Explanation:

Given question is,
${ \red{ \sf{y = {tan}^( - 1)( \frac{3x - {x}^(3) }{1 - 3 {x}^(2) })}}}$

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

${ \tt{ \theta = { \tan}^( - 1) x}}$

${ \red{ \sf{y = { \tan}^( - 1)( \frac{3 \: tan \theta - { \tan}^(3) \theta}{1 - 3 { \tan}^(2) \theta})}}}$

But
${ \green{ \sf{ \tan3 \theta = \frac{3 \: \tan \theta - { \tan}^(3) \theta }{1 - 3 \: { \tan}^(2) \theta}}}}$

${ \red{ \sf{y = { \tan}^( - 1) ( \tan3 \theta)}}}$

${ \red{ \sf{y = 3 \theta}}}$

${ \red{ \sf{y = 3}}}{ \tt{ { \tan}^( - 1) x}}$

${ \red{ \sf{ (dy)/(dx) = 3 * }}}{ \tt{ \frac{1}{ {1 + x}^(2)}}}$

•°•
${ \green{ \boxed{ \red{ \sf{ (dy)/(dx) = \frac{3}{1 + {x}^(2)}}}}}}$

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