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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
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Answer

Consider $\text{y}=\cos(\log\text{ x})^2$
Differentiating it with respect to x and applying the chain and the product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}+\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}$
$=\frac{1}{2}(\text{x}+1)^\frac{-1}{2}+\frac{1}{2}(\text{x}-1)^\frac{-1}{2}$
$=\frac{1}{2}\Big(\frac{1}{\sqrt{\text{x}+1}}\frac{1}{\sqrt{\text{x}-1}}\Big)$
$=\frac{1}{2}\bigg(\frac{\sqrt{\text{x}-1}+\sqrt{\text{x}+1}}{\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big(\frac{\text{y}}{\sqrt{\text{x}^2-1}}\Big)$
So,
$\sqrt{\text{x}^2}-\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}$

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