#### Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 27

Proved

Hint:

You must know the derivative of logarithm and tangent inverse $x$

Given:

$y=\left\{\log \left(x+\sqrt{x^{2}+1}\right)\right\}^{2}, \text { show }\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=2$

Solution:

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

\begin{aligned} &\text { Differentiate the equation w.r.t } x\\ &\frac{d y}{d x}=2 \log \left(x+\sqrt{x^{2}+1}\right) \cdot \frac{1}{x+\sqrt{x^{2}+1}} \times\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)\\ &=\frac{2 \log \left(x+\sqrt{x^{2}+1}\right)}{x+\sqrt{x^{2}+1}} \times \frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\\ &=\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=2 \log \left(x+\sqrt{x^{2}+1}\right) \end{aligned}

\begin{aligned} &\text { Differentiate again }\\ &\frac{d^{2} y}{d x^{2}} \sqrt{1+x^{2}}+\frac{2 x}{2 \sqrt{1+x^{2}}} \frac{d y}{d x}=\frac{2}{x+\sqrt{x^{2}+1}} \times\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)\\ &\frac{\left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}}{\sqrt{x^{2}+1}}=\frac{2}{x+\sqrt{x^{2}+1}} \times \frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\\ &\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=2 \end{aligned}