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If $\mathrm{u}=\mathrm{e}^{\mathrm{ax}} \sin \mathrm{bx}$ and $\mathrm{v}=\mathrm{e}^{\mathrm{ax}} \cos \mathrm{bx}$, then what is $\mathrm{u}$
$\frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}$ equal to?
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$\frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}$ equal to?
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The correct answer is:
a $\mathrm{e}^{2 \mathrm{ax}}$
$\begin{aligned} u & \frac{d u}{d x}+v \frac{d v}{d x}=e^{a x} \sin b x\left[a e^{a x} \sin b x+b e^{a x} \cos b x\right] \\ &+e^{a x} \cos b x\left[a e^{a x} \cos b x-b e^{a x} \sin b x\right] \\=& e^{2 a x}\left[a \sin ^{2} b x+b \sin b x \cos b x+\right.\\ \left.\quad a \cos ^{2} b x-b \sin b x \cos b x\right] \\=& a e^{2 a x} \end{aligned}$
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