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If the function u=e^x \cos x, v=e^x \sin x  then find the equation   v \frac{d u}{d x}-u \frac{d v}{d x} .

Option: 1

v^2+u^2


Option: 2

v^2-u^2


Option: 3

-\left(v^2-u^2\right)


Option: 4

-\left(v^2+u^2\right)


Answers (1)

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Given that, u=e^x \cos x, v=e^x \sin x \begin{aligned} & \frac{d u}{d x}=-e^x \sin x+e^x \cos x=u-v \\ & \frac{d v}{d x}=e^x \cos x+e^x \sin x=u+v \\ & \therefore v \frac{d u}{d x}-u \frac{d v}{d x}=v(u-v)-u(u+v) \\ & =u v-v^2-u^2-u v \\ & =-\left(v^2+u^2\right) \end{aligned}

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Ritika Harsh

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