JEE MAIN - Mathematics (2017 (Offline) - No. 18)
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
$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals
$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$
$${{{3x} \over {1 - 9{x^3}}}}$$
$${{3 \over {1 + 9{x^3}}}}$$
$${{9 \over {1 + 9{x^3}}}}$$
Explanation
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) = $${9 \over {1 + 9{x^3}}}$$
= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$
= 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) = $${9 \over {1 + 9{x^3}}}$$
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