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Question

# If u=tanâˆ’1(x2+y2x+y), then xdudx+ydudy=

A
sin2u
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B
12sin2u
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C
13sin2u
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D
2sin2u
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Solution

## The correct option is C 12sin2u Given, u=tan−1(x2+y2x+y)⇒tanu=x2+y2(x+y)On differentiating both the sides w.r.t. x and y respectively, we getsec2ududx=2x(x+y)−(x2+y2)(x+y)2 ....(1)and sec2ududy=2y(x+y)−(x2+y2)(x+y)2 ....(2)Multiplying x with equation (1) and y to equation (2).Therefore, xsec2ududx=x3+2x2y−xy2(x+y)2 ....(3)and ysec2ududy=y3+2xy2−x2y(x+y)2 ....(4)Adding equations (3) and (4), we getsec2u(xdudx+ydudy)=x3+2x2y−xy2+y3+2xy2−x2y(x+y)2=x3+x2y+y3+xy2(x+y)2=(x+y)(x2+y2)(x+y)2=x2+y2x+y=tanuThus xdudx+ydudy=sinucosu=12sin2u

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