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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\},$ show that $\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{x}^2-1}}.$
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Answer

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

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