Question

If $y=log(\sqrt{x}+\frac{1}{\sqrt{x}}{)}^2,$ then prove that $x(x+1{)}^2{y}_2+(x+1{)}^2{y}_1=2.$

Answer 1

Here $y=log(\sqrt{x}+\frac{1}{\sqrt{x}}{)}^2$ $\Rightarrow y=2log\left(\frac{x+1}{\sqrt{x}}\right)=2log(x+1)-logx$
$\therefore \frac{dy}{dx}=\frac{2}{x+1}-\frac{1}{x}$ $\Rightarrow \frac{dy}{dx}=\frac{x-1}{x(x+1)}...\left(i\right)$
Again differentiating w.r.t. x both sides, we get : $\frac{d^{2}y}{d{x}^2}=\frac{x(x+1)(1-0)-(x-1)(2x+1)}{x^2(x+1{)}^2}$
$x(x+1{)}^2\frac{d^{2}y}{d{x}^2}=\frac{2x-(x-1)(x+1)}{x}=2-\frac{(x-1)}{x(x+1)}\times (x+1{)}^2$
By (i), we get $x(x+1{)}^2\frac{d^{2}y}{d{x}^2}=2-\frac{dy}{dx}\times (x+1{)}^2\therefore x(x+1{)}^2{y}_2+(x+1{)}^2{y}_1=2.$