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Question

If u=tan1(x3+y3xy),

then x22ux2+2xy2uxy+y22uy2=?

A
(2cos2u1)sin2u
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B
(2cos2u+1)sin2u
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C
(sin2u1)cos2u
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D
(sin2u+1)cos2u.
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Solution

The correct option is A (2cos2u1)sin2u
Given : u=tan1(x3+y3xy)

tanu=x3+y3xy

(xy)tanu=x3+y3(1)

Differentiate w.r.t x partially

+tanu+(xy)sec2uux=3x2(2)

Differntiate w.r.t y partially

tanu(xy)sec2uux=3x2(3)

Further x(2) + y(3)

(xy)tan(u)+sec2u(xy){xux+yuy}=3(x3+y3)

(xy)[tan(u)+sec2u{xux+yuy}]=3(x3+y3)

tan(u)+sec2u{xux+yuy}=3(x3+y3)(xy)

tan(u)+sec2u{xux+yuy}=3tanu

xux+yuy=sin(2u)

Differentiating w.r.t x and y gives

ux+x2ux2+x2uxy=2cos(2u)ux(4)

x2uxy+y2uy2+uy=2cos(2u)uy

Further x(4) + y(5) gives

x22ux2+2xy2uxy+y22uy2={2cos(2u)1}{xux+yuy}

={2cos(2u)1}sin(2u)

Hence option A is correct answer

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