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

MCQ

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

✓
${e^x}\left[ {\log (1 + {x^2}) + {{2x} \over {1 + {x^2}}}} \right]$
B
${e^x}\left[ {\log (1 + {x^2}) - {{2x} \over {1 + {x^2}}}} \right]$
C
${e^x}\left[ {\log (1 + {x^2}) + {x \over {1 + {x^2}}}} \right]$
D
${e^x}\left[ {\log (1 + {x^2}) - {x \over {1 + {x^2}}}} \right]$

✓

Answer

Correct option: A.

${e^x}\left[ {\log (1 + {x^2}) + {{2x} \over {1 + {x^2}}}} \right]$

a
(a) $\frac{d}{{dx}}\{ {e^x}\log (1 + {x^2})\} = {e^x}\log (1 + {x^2}) + {e^x}\frac{1}{{(1 + {x^2})}}2x$

$ = {e^x}\left[ {\log (1 + {x^2}) + \frac{{2x}}{{1 + {x^2}}}} \right]$.

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