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

MCQ

${d \over {dx}}\left( {{{{e^x}} \over {1 + {x^2}}}} \right) = $

A
${{{e^x}(1 + x)} \over {{{(1 + {x^2})}^2}}}$
✓
${{{e^x}{{(1 - x)}^2}} \over {{{(1 + {x^2})}^2}}}$
C
${{{e^x}{{(1 + x)}^2}} \over {(1 + {x^2})}}$
D
${{{e^x}{{(1 - x)}^2}} \over {(1 + {x^2})}}$

✓

Answer

Correct option: B.

${{{e^x}{{(1 - x)}^2}} \over {{{(1 + {x^2})}^2}}}$

b
(b) $\frac{d}{{dx}}\left( {\frac{{{e^x}}}{{1 + {x^2}}}} \right) = \frac{{(1 + {x^2}){e^x} - {e^x}(2x)}}{{{{(1 + {x^2})}^2}}} = \frac{{{e^x}{{(1 - x)}^2}}}{{{{(1 + {x^2})}^2}}}$

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