Question

If \[ y=\left(x+\sqrt{x^{2}+1}\right)^{5}, \] then \(25y=\)

• A. \((x^{2}+1)y_{2}-xy_{1}\)
• B. \((x^{2}+1)y_{2}+xy_{1}\)
• C. \((x^{2}+1)y_{2}-2xy_{1}\)
• D. \((x^{2}+1)y_{2}+2xy_{1}\)

Hint:

Remember the standard identity \[ \boxed{ \frac{d}{dx} \ln\!\left(x+\sqrt{x^{2}+1}\right) = \frac1{\sqrt{x^{2}+1}}. } \] It greatly simplifies repeated differentiation problems involving \[ x+\sqrt{x^{2}+1}. \]

Answer 1

The Correct Option is B

Step 1: Take logarithm and differentiate. Given, \[ y=\left(x+\sqrt{x^{2}+1}\right)^5. \] Taking logarithm, \[ \ln y = 5\ln\!\left(x+\sqrt{x^{2}+1}\right). \] Since \[ \frac{d}{dx} \ln\!\left(x+\sqrt{x^{2}+1}\right) = \frac1{\sqrt{x^{2}+1}}, \] we obtain \[ \frac{y'}{y} = \frac5{\sqrt{x^{2}+1}}. \] Hence, \[ y' = \frac{5y}{\sqrt{x^{2}+1}}. \]

Step 2: Differentiate once again. Differentiating, \[ y'' = \frac{5y'}{\sqrt{x^{2}+1}} - \frac{5xy}{(x^{2}+1)^{3/2}}. \] Substituting \[ y' = \frac{5y}{\sqrt{x^{2}+1}}, \] gives \[ y'' = \frac{25y}{x^{2}+1} - \frac{5xy}{(x^{2}+1)^{3/2}}. \] Multiplying throughout by \[ x^{2}+1, \] \[ (x^{2}+1)y'' = 25y - \frac{5xy}{\sqrt{x^{2}+1}}. \] Using \[ y' = \frac{5y}{\sqrt{x^{2}+1}}, \] we get \[ \boxed{ 25y = (x^{2}+1)y'' + xy'. } \] Using the notation \[ y_1=y', \qquad y_2=y'', \] we obtain \[ \boxed{ 25y=(x^{2}+1)y_{2}+xy_{1}. } \] Hence, the correct option is \(\boxed{(B)}\).