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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=(\text{x}-1)\log(\text{x}-1)-(\text{x}+1)\log(\text{x}+1)$ prove that $\frac{\text{dy}}{\text{dx}}=\log\Big(\frac{\text{x}-1}{1+\text{x}}\Big)$
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Answer

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

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