#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 45

Answer: $2 x+\frac{2 e^{x}}{\sqrt{x^{4}-1}}$

Hint: you must know the rules of solving derivative of polynomials

Given: $\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}$

Solution:

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

By rationalizing,

$\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}} \times \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}$

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

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

$=\frac{x^{2}+1+x^{2}-1+2 \sqrt{x^{4}-1}}{2}$

$=x^{2}+\sqrt{x^{4}-1}$

Now, let  $y=x^{2}+\sqrt{x^{4}-1}$

Now, differentiate

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+\sqrt{x^{4}-1}\right)$

\begin{aligned} &\Rightarrow 2 x+\frac{1}{2 \sqrt{x^{4}-1}} \times\left(4 x^{3}\right) \\\\ &\Rightarrow 2 x+\frac{2 x^{3}}{\sqrt{x^{4}-1}} \end{aligned}