MCQ
${d \over {dx}}\{ {e^{ - a{x^2}}}\log (\sin x)\} = $
- A${e^{ - a{x^2}}}(\cot x + 2ax\log \sin x)$
- B${e^{ - a{x^2}}}(\cot x + ax\log \sin x)$
- ✓${e^{ - a{x^2}}}(\cot x - 2ax\log \sin x)$
- DNone of these
✓
Answer
Correct option: C.
${e^{ - a{x^2}}}(\cot x - 2ax\log \sin x)$
c
(c) $\frac{d}{{dx}}\{ {e^{ - a{x^2}}}\log (\sin x)\} $
(c) $\frac{d}{{dx}}\{ {e^{ - a{x^2}}}\log (\sin x)\} $
$ = {e^{ - a{x^2}}}( - 2ax).\log (\sin x) + {e^{ - a{x^2}}}\cot x$
$ = {e^{ - ax}}^2[\cot x - 2ax\log (\sin x)]$.
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