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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
If $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
✓

Answer

We have, $\text{y}=\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}}\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{1}{2}\frac{\sqrt{\text{x}}}{\text{x}+1}\Big(\frac{\text{x}-1}{\text{x}\sqrt{\text{x}}}\Big)$
$=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{2\text{x}(\text{x}+1)}$

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