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$\frac{d}{d x} \{e^{a x} \sin (b x + c)\} =$

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$e^{a x} [a \sin (b x + c) + b \cos (b x + c)]$

$e^{a x} [a \cos (b x + c) + b \sin (b x + c)]$

$e^{a x} [a \sin (b x + c) - b \cos (b x + c)]$

$e^{a x} [a \cos (b x + c) - b \sin (b x + c)]$

answer is A.

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Detailed Solution

$\frac{d}{d x} {e^{a x} \sin (b x + c)}$

$= e^{a x} . \cos (b x + c) . b + \sin (b x + c) . e^{a x} . a$

$= e^{a x} (b \cos (b x + c) + a \sin (b x + c))$

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