Question

$If{y}^{1/n}=\{x+\sqrt{(1+x^2)}\},$ $then(1+x^2)y_2+x{y}_{1}isequalto$

• A. $n^{2}y$
• B. $n{y}^2$
• C. $n^2{y}^2$
• D. $Noneofthese$

Answer 1

The correct option is A $n^{2}y$

$Since{y}^{1/n}=\{x+\sqrt{(1+x^2)}\}$
$ory=(x+\sqrt{1+x^2}{)}^n$
$\therefore y_1=(x+\sqrt{1+x^2}{)}^{n-1}(1+\frac{x}{\sqrt{1+x^2}})$
$\sqrt{(1+x^2)}{y}_1=n(x+\sqrt{1+x^2}{)}^n$
$\sqrt{(1+x^2)}.y_1=ny$
Squaring both sides
$\therefore (1+x^2)y_1^2=n^2{y}^2$
Differentiating both sides w.r.t. x
$\therefore (1+x^2{)}^2{y}_1{y}_2+y_1^2.2x=2n^{2}y{y}_1$
$2y_1\ne 0$
$\therefore (1+x^2)y_2+x{y}_1=n^{2}y$