d dx [ e^ ax (bx + c) ] = - Vidyadip

MCQ

${d \over {dx}}\left[ {{{{e^{ax}}} \over {\sin (bx + c)}}} \right] = $

A
${{{e^{ax}}[a\sin (bx + c) + b\cos (bx + c)]} \over {{{\sin }^2}(bx + c)}}$
B
${{{e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]} \over {\sin (bx + c)}}$
✓
${{{e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]} \over {{{\sin }^2}(bx + c)}}$
D
None of these

✓

Answer

Correct option: C.

${{{e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]} \over {{{\sin }^2}(bx + c)}}$

c
(c) $\frac{d}{{dx}}\left( {\frac{{{e^{ax}}}}{{\sin (bx + c)}}} \right)$$ = \frac{{a{e^{ax}}\sin (bx + c) - b{e^{ax}}\cos (bx + c)}}{{{{\{ \sin (bx + c)\} }^2}}}$

$ = \frac{{{e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]}}{{{{\sin }^2}(bx + c)}}$.

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