Let y=f(x) is a positive function which satisfies equation √y2+2x+√y2−2x=2x2,then dydx is
A
2x3−x−3y
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B
2x4−x−2y
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C
2x4−x−2√x6+1
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D
2x3−x−3√1+x6
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Solution
The correct options are A2x3−x−3y C2x4−x−2√x6+1 √y2+2x+√y2−2x=2x2...(1) rationalizing the equation (1) we get √y2+2x−√y2−2x=2x ...(2) adding the equations (1) & (2) 2√y2+2x=2x2+2x (y2+2x)=(x2+1x)2 y2=x4+1x2⇒y=√x4+1x2 2y.dydx=4x3−2x−3 dydx=2x3−x−3y dydx=2x3−x−3√1+x6x2=2x4−x−2√1+x6