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Mathematics Question on Statistics

If xy+yx=a\sqrt{\frac{x}{y}} + \sqrt{ \frac{y}{x}} = \sqrt{a} , then dydx=\frac{dy}{dx} =

A

xy\frac{x}{y}

B

yx\frac{y}{x}

C

xx

D

0

Answer

yx\frac{y}{x}

Explanation

Solution

xy+yx=a\sqrt{\frac{x}{y}} + \sqrt{ \frac{y}{x}} = \sqrt{a} ....(i)
Squaring both sides of (i), we get
xy+yx+2=a\frac{x}{y} + \frac{y}{x} + 2 = a
or x2+y2=(a−2)xyx^{2} + y^{2} =\left(a-2\right)xy .....(ii)
Differentiating w.r.t. x, we have
2x+2ydydx=(a−2)(y+xdydx)2x+2y \frac{dy}{dx} =\left(a-2\right) \left(y+x \frac{dy}{dx}\right)
⇒2ydydx−(a−2)xdydx=(a−2)y−2x\Rightarrow 2y \frac{dy}{dx} -\left(a -2\right)x \frac{dy}{dx}=\left(a-2\right)y -2x
⇒dydx[2y−(a−2)x]=(a−2)y−2x\Rightarrow \frac{dy}{dx} \left[2y -\left(a -2\right)x\right] =\left(a-2\right)y-2x
⇒dydx=(a−2)y−2x2y−(a−2)x\Rightarrow \frac{dy}{dx} =\frac{\left(a -2\right)y-2x}{2y-\left(a-2\right)x}
⇒dydx=[(x2+y2xy)y−2x2y−(x2+y2xy)x]\Rightarrow \frac{dy}{dx} = \left[\frac{\left(\frac{x^{2} +y^{2}}{xy}\right)y - 2x}{2y -\left(\frac{x^{2} +y^{2}}{xy}\right)x}\right] (From (ii))
=x2+y2−2x2x(2y2−x2−y2y)=yx(y2−x2y2−x2)=yx= \frac{x^{2} +y^{2}-2x^{2}}{x\left(\frac{2y^{2} -x^{2} -y^{2}}{y}\right)} = \frac{y}{x} \left(\frac{y^{2}-x^{2}}{y^{2}-x^{2}}\right) =\frac{y}{x}