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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big),$ prove that $\big(\text{x}^2+1\big)\frac{\text{dx}}{\text{dx}}+\text{xy}+1=0$
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Answer

We have, $\text{y}\sqrt{\text{x}^2+1}=\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$Differentiating with respect to x, we get,
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{y}\sqrt{\text{x}^2+1}\Big)=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}^2+1}-\text{x}\Big)$
[Using Product rule and chain rule]
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}\big)+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$\Rightarrow\frac{\text{y}}{2\sqrt{\text{x}^2+1}}\times\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\sqrt{\text{x}^2+1}\frac{\text{d}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\times\Big[\frac{1}{2\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-1\Big]$
$\Rightarrow\ \frac{2\text{xy}}{2\sqrt{\text{x}^2+1}}+\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\frac{1}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}\Big[\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}-1\Big]$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}} \\ =\Big[\frac{1}{\sqrt{\text{x}^2+1}-\text{x}}\Big]\Big[\frac{\text{x}-\sqrt{\text{x}^2+1}}{\sqrt{\text{x}^2+1}}\Big]-\frac{\text{xy}}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\sqrt{\text{x}^2+1}\frac{\text{dy}}{\text{dx}}=\frac{-(1+\text{xy})}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\big(\text{x}^2+1\big)\frac{\text{dy}}{\text{dx}}=-(1+\text{xy})$
$\Rightarrow(\text{x}^2+1)\frac{\text{dy}}{\text{dx}}+1+\text{xy}=0$

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