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Mathematics Question on Statistics

If xx and yy are related by the relationx2−ax+y2=0x^2 - ax+ y^2 = 0, then d2y/dx2d^2y/dx^2 is equal to

A

a2(4y3)\frac{a^2}{(4y^3)}

B

−a2(4y3)\frac{-a^2}{(4y^3)}

C

(4y2−a2(4y3)\frac{(4y^2 - a^2}{(4y^3)}

D

(4y2+a2(4y3)\frac{(4y^2 + a^2}{(4y^3)}

Answer

−a2(4y3)\frac{-a^2}{(4y^3)}

Explanation

Solution

x2−ax+y2=0⇒ y2=ax−x2       x^2 - ax+ y^2 = 0 \Rightarrow \, y^2 = ax - x^2 \, \, \, \, \, \, \, .....(i)
Differentiating w.r.t. 'x', we get
2ydydx=a−2x ⇒  dydx=a−2x2y     2 y \frac{dy}{dx} = a - 2x \, \Rightarrow \, \, \frac{dy}{dx} = \frac{a -2x}{2y} \, \, \, \, \, ...(ii)
Differentiating again w.r.t. 'x', we get
d2ydx2=2y(−2)−(a−2x)2dydx4y2\frac{d^{2}y}{dx^{2}} = \frac{2y\left(-2\right) -\left(a -2x\right)2 \frac{dy}{dx}}{4y^{2}}
=−4y−2(a−2x)(a−2x)2y4y2= \frac{-4y - 2\left(a-2x\right) \frac{\left(a -2x\right)}{2y}}{4y^{2}} (from (ii))
=−4y2−(a−2x)24y3= \frac{-4y^{2} -\left(a -2x\right)^{2}}{4y^{3}}
=−4(ax−x2)−(a−2x)24y3=\frac{-4\left(ax -x^{2}\right) -\left(a-2x\right)^{2}}{4y^{3}} (from (i))
=−a24y3=\frac{-a^{2}}{4y^{3}}