If y = x log(xa+bx) then x3d2ydx2 is equal to

On solving we have yx=log x - log(a+bx)⇒xdydx-yx2=1x - ba+bx⇒xdydx- y=axa+bx⇒xd2ydx2+dydx - dydx=a(a+bx)- abx(a+bx)2⇒x3d2ydx2=a2x2(a+bx)2=(xdydx- y)2

If y = x log(xa+bx) then x3d2ydx2 is equal to

On solving we have yx=log x - log(a+bx)⇒xdydx-yx2=1x - ba+bx⇒xdydx- y=axa+bx⇒xd2ydx2+dydx - dydx=a(a+bx)- abx(a+bx)2⇒x3d2ydx2=a2x2(a+bx)2=(xdydx- y)2

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If $y = x log (\frac{x}{a+bx})$ then $x^{3} \frac{d^{2} y}{{dx}^{2}}$ is equal to

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${(x \frac{dy}{dx} -y)}^{2}$

$\frac{a^{2} x}{{(a+bx)}^{2}}$

${(\frac{dy}{dx} -y)}^{2}$

${(x \frac{dy}{dx} +y)}^{2}$

answer is A.

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Detailed Solution

On solving we have $\frac{y}{x} =log x - log (a+bx)$
$\Rightarrow \frac{x \frac{dy}{dx} -y}{x^{2}} = \frac{1}{x} - \frac{b}{a+bx}$

$\Rightarrow x \frac{dy}{dx} - y= \frac{ax}{a+bx} \Rightarrow x \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{a (a+bx) - abx}{{(a+bx)}^{2}}$
$\Rightarrow x^{3} \frac{d^{2} y}{{dx}^{2}} = \frac{a^{2} x^{2}}{{(a+bx)}^{2}} = {(x \frac{dy}{dx} - y)}^{2}$