If y=x22+12xx2+1+logex+x2+1, then

y=x22+12xx2+1+12logex+x2+1⇒y'=x+12x2x2+1+x2+1+12x2+1 =x+12x2+1x2+1+x2+12 =x+x2+1Also, 2y=x2+xx2+1+logex+x2+1 =xx+x2+1+logex+x2+1 =xy'+logey'

If y=x22+12xx2+1+logex+x2+1, then

y=x22+12xx2+1+12logex+x2+1⇒y'=x+12x2x2+1+x2+1+12x2+1 =x+12x2+1x2+1+x2+12 =x+x2+1Also, 2y=x2+xx2+1+logex+x2+1 =xx+x2+1+logex+x2+1 =xy'+logey'

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$If y = \frac{x^{2}}{2} + \frac{1}{2} x \sqrt{x^{2} + 1} + \log_{e} \sqrt{x + \sqrt{x^{2} + 1}}, then$

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$y' = x + \sqrt{x^{2} + 1}$

$2 y = xy' + \log_{e} y'$

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

$2 y = xy' - \log_{e} y'$

answer is A, C.

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Detailed Solution

$\begin{array}{l} y = \frac{x^{2}}{2} + \frac{1}{2} x \sqrt{x^{2} + 1} + \frac{1}{2} \log_{e} (x + \sqrt{x^{2} + 1}) \\ \Rightarrow y' = x + \frac{1}{2} [\frac{x^{2}}{\sqrt{x^{2} + 1}} + \sqrt{x^{2} + 1}] + \frac{1}{2 \sqrt{x^{2} + 1}} \\ = x + \frac{1}{2} [\frac{x^{2} + 1}{\sqrt{x^{2} + 1}}] + \frac{\sqrt{x^{2} + 1}}{2} \\ = x + \sqrt{x^{2} + 1} \\ Also, 2 y = x^{2} + x \sqrt{x^{2} + 1} + \log_{e} (x + \sqrt{x^{2} + 1}) \\ = x (x + \sqrt{x^{2} + 1}) + \log_{e} (x + \sqrt{x^{2} + 1}) \\ = xy' + \log_{e} y' \end{array}$