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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\Big(\text{x}^{-1\frac{1}{2}}\Big)$
$=\frac{1}{2\sqrt{\text{x}}}+\Big(-\frac{1}{2}\times\text{x}^{-\frac{1}{2}-1}\Big)$
$=\frac{2}{2\sqrt{\text{x}}}-\frac{1}{2\sqrt[\text{x}]{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}\sqrt{\text{x}}}$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$
Hence, the solution is, $2\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}}$

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