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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate the function given in Exercise:
$\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}$
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Answer

Let $\text{y}=\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}=\Big(\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}\Big)^{\frac{1}{2}}\ \dots\text{(i)}$
Taking logs on both sides, we have
$\log\text{y}=\frac{1}{2}[\log(\text{x}-1)+\log(\text{x}-2)-\log(\text{x}-3)-\log(\text{x}-4)-\log(\text{x}-5)]$
$\therefore\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{1}{\text{x}-1}\frac{\text{d}}{\text{dx}}(\text{x}-1)+\frac{1}{\text{x}-2}\frac{\text{d}}{\text{dx}}(\text{x}-2)-\frac{1}{\text{x}-3}\frac{\text{d}}{\text{dx}}(\text{x}-3)-\frac{1}{\text{x}-4}\frac{\text{d}}{\text{dx}}(\text{x}-4)-\frac{1}{\text{x}-5}\frac{\text{d}}{\text{dx}}(\text{x}-5)\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}\Big[\frac{1}{\text{x}-1}+\frac{1}{\text{x}-2}-\frac{1}{\text{x}-3}-\frac{1}{\text{x}-4}-\frac{1}{\text{x}-5}\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\sqrt{\frac{(\text{x}-1)(\text{x}-2)}{(\text{x}-3)(\text{x}-4)(\text{x}-5)}}\Big[\frac{1}{\text{x}-1}+\frac{1}{\text{x}-2}-\frac{1}{\text{x}-3}-\frac{1}{\text{x}-4}-\frac{1}{\text{x}-5}\Big]\ \text{[From eq.(i)}]$

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