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Question

If xexy=y+sin2x, then at x=0, dydx=?

A
1
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B
2
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C
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D
0
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Solution

The correct option is B 1
Given
xexy=y+sin2x
Differentiate on both sides
exy+xexy(1+dydx)=dydx+2sinxcosx
at x=0
1+0×e0y(1+dydx)=dydx+0
dydx=1.

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