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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
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Answer

We have, $\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}$
By rationalising we get,
$\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}}\times\frac{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$
$=\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)^2}{\big(\sqrt{\text{x}^2+1}\big)^2-\big(\sqrt{\text{x}^2-1}\big)^2}$
$=\frac{\big(\sqrt{\text{x}^2+1}\big)^2+\big(\sqrt{\text{x}^2-1}\big)^2+2\big(\sqrt{\text{x}^2+1}\big)\big(\sqrt{\text{x}^2-1}\big)}{\text{x}^2+1-\text{x}^2+1}$
$=\frac{\text{x}^2+1+\text{x}^2-1+2\sqrt{\text{x}^4-1}}{2}$
$=\frac{2\text{x}^2+2\sqrt{\text{x}^4-1}}{2}$
$=\text{x}^2+\sqrt{\text{x}^4-1}$
Now, Let $\text{y}=\text{x}^2+\sqrt{\text{x}^4-1}$
Differentiate it with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\sqrt{\text{x}^4-1}\big)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^4-1)$
$=2\text{x}+\frac{1}{2\sqrt{\text{x}^4-1}}\times(4\text{x}^3)$
$=2\text{x}+\frac{2\text{x}^3}{\sqrt{\text{x}^4-1}}$

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