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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
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Answer

Here,
$\sqrt{\text{y}+\text{x}}+\sqrt{\text{y}-\text{x}}=\text{c}$
Differentiating with respect to x,
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sqrt{\text{y}+\text{x}})+\frac{\text{d}}{\text{dx}}\sqrt{\text{y}-\text{x}}=\frac{\text{d}}{\text{dx}}(\text{c})$
$\Rightarrow\frac{1}{2\sqrt{\text{y}+\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}+\text{x})+\frac{1}{2\sqrt{\text{y}-\text{x}}}\frac{\text{d}}{\text{dx}}(\text{y}-\text{z})=0$
$\Rightarrow \frac{1}{2\sqrt{\text{y}+\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}+1\Big)+\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}+\text{x}}}\Big)+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{2\sqrt{\text{y}-\text{x}}}\Big)=\frac{1}{2\sqrt{\text{y}-\text{x}}}-\frac{1}{2\sqrt{\text{y}+\text{x}}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\times\Big[\frac{1}{\sqrt{\text{y}+\text{x}}}+\frac{1}{\sqrt{\text{y}-\text{x}}}\Big]=\frac{1}{2}\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big[\frac{\sqrt{\text{y}-\text{x}}-\sqrt{\text{y}+\text{x}}}{\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}\Big]=\Big[\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}\sqrt{\text{y}+\text{x}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}-\text{x}}+\sqrt{\text{y}-\text{x}}}\times\frac{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}-\text{x}}}{\sqrt{\text{y}+\text{x}}-\sqrt{\text{y}+\text{x}}}$
[Rationalizing the denominator]
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}+\text{x})+(\text{y}-\text{x})-2\sqrt{\text{y}+\text{x}}\sqrt{\text{y}-\text{x}}}{\text{y}+\text{x}-\text{y}+\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}-2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{2\text{x}}-\frac{2\sqrt{\text{y}^2-\text{x}^2}}{2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2-\text{x}^2}{\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$

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