Solution
Given that
y=u2−12u...(1)u=x2...(2)
From (1),
y=u2−12ududy=(u2−1)2(u2−1)dud(2u)−(2u)dud(u2−1)dudy=(u2−1)2(u2−1)(2)−(2u)(2u)dudy=(u2−1)2(2u2−2)−(4u2)dudy=(u2−1)2(−2u2−2)dudy=(u2−1)2−2(u2+1)...(3)
Now from (2)
u=x2dxdu=dxd(x2)dxdu=2x...(4)
According to the chain rule
dxdy=dudy×dxdu
Using (3) and (4)
dxdy=dudy×dxdudxdy=(u2−1)2−2(u2+1)×(2x)dxdy=(u2−1)2−4x(u2+1)
Hence according to the chain rule,
dxdy=(u2−1)2−4x(u2+1)
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