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Mathematics Question on Statistics

If y=x+y+x+y+.....y =\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y}}}} +..... then dydx\frac{dy}{dx}

A

y+xy2−2x\frac{y+x}{y^{2} -2x}

B

y3−x2y2−2xy−1\frac{y^3 - x}{2y^{2} -2xy - 1}

C

y3+x2y2−x\frac{y^3 +x}{2y^{2} - x}

D

y2−x2y3−2xy−1\frac{y^2 - x}{2y^{3} -2xy - 1 }

Answer

y2−x2y3−2xy−1\frac{y^2 - x}{2y^{3} -2xy - 1 }

Explanation

Solution

y=x+y+x+y+....y =\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y}}}}+....
then, y can be rewritten as,
y=x+y+y=x+2yy= \sqrt{x+\sqrt{\sqrt{y+y}}} = \sqrt{x+\sqrt{2y}} ......(i)
Squaring (i) both sides, we get
y2=x+2y⇒y2−x=2yy^{2} =x+\sqrt{2y} \Rightarrow y^{2} -x =\sqrt{2y} ...... (ii)
Again squaring (ii) both sides, we get
y4+x2−2xy2=2yy^{4}+x^{2}-2x y^{2} =2y ........(iii)
Differentiating (iii) w.r.t. 'x', we have
⇒4y3dydx+2x−2(y2+x.2ydydx)=2dydx\Rightarrow 4y^{3} \frac{dy}{dx} + 2x - 2 \left(y^{2} +x.2y \frac{dy}{dx}\right) = \frac{2dy}{dx}
⇒4y3dydx−4xydydx=2dydx=−2x+2y2\Rightarrow 4y^{3} \frac{dy}{dx} - 4xy \frac{dy}{dx} = \frac{2dy}{dx} = - 2x + 2y^{2}
⇒(2y3−2xy−1)dydx=y2−x\Rightarrow \left(2y^{3} - 2xy -1\right) \frac{dy}{dx}=y^{2} -x
⇒dydx=y2−x2y3−2xy−1\Rightarrow \frac{dy}{dx}= \frac{y^{2}-x}{2y^{3} -2xy -1}