Question

If $y=f\left(x\right)=(x+\sqrt{1+x^2}{)}^n$, then $(1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}$ equals

• A. $f\left(x\right)$
• B. $0$
• C. $n^{2}f\left(x\right)$
• D. $nf\left(x\right)$

Answer 1

The correct option is C $n^{2}f\left(x\right)$

Given equation: $y=f\left(x\right)={(x+\sqrt{1+x^2})}^n$

$y^{{}^{\prime }}=n{(x+\sqrt{1+x^2})}^{n-1}(1+\frac{2x}{2\sqrt{1+x^2}})$

$=n{(x+\sqrt{1+x^2})}^{n-1}\left(\frac{(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\right)$

$=\frac{n{(x+\sqrt{1+x^2})}^n}{\sqrt{1+x^2}}$

$=\frac{ny}{\sqrt{1+x^2}}-------\left(1\right)$

$y^{{}^{\prime \prime }}=\frac{n{y}^{{}^{\prime }}}{\sqrt{1+x^2}}-\frac{nyx}{(1+x^2{)}^{\frac{3}{2}}}-------\left(2\right)$

now putting these values in the equation, we get

$=(1+x^2)y^{{}^{\prime \prime }}+x{y}^{{}^{\prime }}=(1+x^2)(\frac{n{y}^{{}^{\prime }}}{\sqrt{1+x^2}}-\frac{nyx}{{(1+x^2)}^{\frac{3}{2}}})+x\left(\frac{ny}{\sqrt{1+x^2}}\right)$

$=\left(\sqrt{1+x^2}\right)n{y}^{{}^{\prime }}-\frac{nyx}{\sqrt{1+x^2}}+\frac{nyx}{\sqrt{1+x^2}}$

$=\left(\sqrt{1+x^2}\right)ny=n^{2}y=n^{2}f\left(x\right)$

So, option (c) is correct.