d dx ^ - 1 ( ax - b bx + a ) = - Vidyadip

MCQ

${d \over {dx}}{\tan ^{ - 1}}\left( {{{ax - b} \over {bx + a}}} \right) = $

A
${1 \over {1 + {x^2}}} - {{{a^2}} \over {{a^2} + {b^2}}}$
B
${{ - 1} \over {1 + {x^2}}} - {{{a^2}} \over {{a^2} + {b^2}}}$
C
${1 \over {1 + {x^2}}} + {{{a^2}} \over {{a^2} + {b^2}}}$
✓
None of these

✓

Answer

Correct option: D.

None of these

d
(d) $\frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{ax - b}}{{bx + a}}} \right)= \frac{1}{{1 + {{\left( {\frac{{ax - b}}{{bx + a}}} \right)}^2}}}.$$\frac{d}{{dx}}\left( {\frac{{ax - b}}{{bx + a}}} \right)$

$ = \frac{{{a^2} + {b^2}}}{{{a^2} + {b^2} + {a^2}{x^2} + {b^2}{x^2}}} = \frac{1}{{1 + {x^2}}}$.

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