Question
Differentiate the following functions with respect to x:
$\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
$\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
✓
Answer
Let, $\text{y}=\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\Big]$
$=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)$
[Using chain rule and quotient rule]
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{\text{x}^2-\text{x}+1\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^3-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+\text{x}+1)(\text{x}^2+\text{x}+1)}$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+1)^2-(\text{x})^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+1+2\text{x}^2-\text{x}^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\Big\}=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\Big]$
$=\frac{1}{\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)$
[Using chain rule and quotient rule]
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{\text{x}^2-\text{x}+1\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)-(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{(\text{x}^2-\text{x}+1)(2\text{x}+1)-(\text{x}^2+\text{x}+1)(2\text{x}-1)}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^3-\text{x}+1}\Big)\bigg[\frac{2\text{x}^3-2\text{x}^2+2\text{x}+\text{x}^3-\text{x}+1-2\text{x}^3-2\text{x}^2-2\text{x}+\text{x}^2+\text{x}+1}{(\text{x}^2-\text{x}+1)^2}\bigg]$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+\text{x}+1)(\text{x}^2+\text{x}+1)}$
$=\frac{-4\text{x}^2+2\text{x}^3+2}{(\text{x}^2+1)^2-(\text{x})^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+1+2\text{x}^2-\text{x}^2}$
$=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
So,
$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big)\Big\}=\frac{-2(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}$
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