If for [x ( 0,1 4 )], the derivatives of [ ^ - 1( 6x x 1 - 9x^3 )] is [ x .g( x )], then […

Question

If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals

Accepted Answer

Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$

= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$

= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$

$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$

= $${9 \over {1 + 9{x^3}}}.\sqrt x $$

$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$

Solutions · 1