d dx ^ - 1 ( x 1 + 6 x^2 ) = ...... - Vidyadip

MCQ

$\frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{x}{{1 + 6{x^2}}}} \right) = ......$

A
$\frac{{{{\left( {1 + 6{x^2}} \right)}^2}}}{{1 + 7{x^2}}}$
B
$\frac{1}{{1 + 9{x^2}}} - \frac{1}{{1 + 4{x^2}}}$
C
$\frac{3}{{1 + 9{x^2}}} + \frac{2}{{1 + 4{x^2}}}$
✓
$\frac{3}{{1 + 9{x^2}}} - \frac{2}{{1 + 4{x^2}}}$

✓

Answer

Correct option: D.

$\frac{3}{{1 + 9{x^2}}} - \frac{2}{{1 + 4{x^2}}}$

D

$\frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{x}{{1 + 6{x^2}}}} \right)$

$ = \frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{3x - 2x}}{{1 + 3x.2x}}} \right)$

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

$ = \frac{1}{{1 + 9{x^2}}}\frac{d}{{dx}}3x - \frac{1}{{1 + 4{x^2}}}\frac{d}{{dx}}2x$

$ = \frac{3}{{1 + 9{x^2}}} - \frac{2}{{1 + 4{x^2}}}$

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