The correct option is
D sin2uThe given information is:
u=tan−1(3x2y+2xy2x+y)
Taking partial differentiation w.r.t. to x and y one at a time we get,
⇒∂u∂x=11+x2y2(3x+2y)2(x+y)2.(6xy+2y2)(x+y)−(3x2y+2xy2)(x+y)2
⇒∂u∂x=(6xy+2y2)(x+y)−(3x2y+2xy2)(x+y)2+x2y2(3x+2y)2
⇒∂u∂x=6x2y+6xy2+2xy2+2y3−3x2y−2xy2(x+y)2+x2y2(3x+2y)2
⇒∂u∂x=3x2y+6xy2+2y3(x+y)2+x2y2(3x+2y)2
⇒∂u∂y=11+x2y2(3x+2y)2(x+y)2.(6xy+3x2)(x+y)−(3x2y+2xy2)(x+y)2
⇒∂u∂y=(6xy+3x2)(x+y)−(3x2y+2xy2)(x+y)2+x2y2(3x+2y)2
⇒∂u∂y=3x3+3x2y+6x2y+6xy2−3x2y−2xy2(x+y)2+x2y2(3x+2y)2
⇒∂u∂y=3x3+6x2y+4xy2(x+y)2+x2y2(3x+2y)2
Now we make calculation as follows,
⇒x∂u∂x=3x3y+6x2y2+2xy3(x+y)2+x2y2(3x+2y)2
⇒y∂u∂y=3x3y+6x2y2+4xy3(x+y)2+x2y2(3x+2y)2
⇒x∂u∂x+y∂u∂y=6x3y+6xy3+12x2y2(x+y)2+x2y2(3x+2y)2
Now, tanu=3x2y+2xy2x+y
⇒sinu=3x2y+2xy2√(x+y)2+(3x2y+2xy2)2
⇒cosu=x+y√(x+y)2+(3x2y+2xy2)2
⇒sin2u=2sinucosu
⇒sin2u=2(x+y)(3x2y+2xy2)(x+y)2+(3x2y+2xy2)2
⇒sin2u=6x3y+6xy3+12x2y2(x+y)2+x2y2(3x+2y)2
So, ⇒x∂u∂x+y∂u∂y=sin2u