Question

If $y=\frac{x^2}{2}+\frac{1}{2}\times \sqrt{x^2+1}$, then $2y=x{y}^{\prime }$, where $y^{\prime }$ denotes the derivative of $y$ w.r.t. $x$.

• A. True
• B. False

Answer 1

The correct option is B False

Given $y=\frac{x^2}{2}+\frac{1}{2}\times \sqrt{x^2+1}$

$y^{\prime }=\frac{dy}{dx}=\frac{2x}{2}+\frac{1}{2}\times \frac{1}{2}\frac{2x}{\sqrt{x^2+1}}=x+\frac{x}{2\sqrt{x^2+1}}$

Now,

$2y=x{y}^{\prime }⟹x^2+\sqrt{x^2+1}=x^2+\frac{x^2}{\sqrt{x^2+1}}$; which is not true.

Hence given statement is false.