Question
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
✓
Answer
Let $\text{y}=\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
$\Rightarrow\text{y}=\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\big[\log(\text{x}-1)-\log(\text{x}+1)\big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}-1)\big\}-\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}+1)\big\}\Big]$
$=\frac{1}{2}\Big(\frac{1}{\text{x}-1}-\frac{1}{\text{x}+1}\Big)$
$=\frac{1}{2}\Big(\frac{2}{\text{x}^2-1}\Big)$
$=\frac{2}{\text{x}^2-1}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}^2-1}$
$\Rightarrow\text{y}=\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log\Big(\frac{\text{x}-1}{\text{x}+1}\Big)$
$\Rightarrow\text{y}=\frac{1}{2}\big[\log(\text{x}-1)-\log(\text{x}+1)\big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\Big[\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}-1)\big\}-\frac{\text{d}}{\text{dx}}\big\{\log(\text{x}+1)\big\}\Big]$
$=\frac{1}{2}\Big(\frac{1}{\text{x}-1}-\frac{1}{\text{x}+1}\Big)$
$=\frac{1}{2}\Big(\frac{2}{\text{x}^2-1}\Big)$
$=\frac{2}{\text{x}^2-1}$
So,
$\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{x}^2-1}$
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