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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
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Answer

Let $\text{y}=\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big]$
$\frac{\text{d}}{\text{dx}}\log(3\text{x}+2)-\frac{\text{d}}{\text{dx}}\big(\text{x}^2\log(2\text{x}-1)\big)$
$=\frac{1}{3\text{x}+2}\frac{\text{d}}{\text{dx}}(3\text{x}+2)-\Big[\text{x}^2\frac{\text{d}}{\text{dx}}\log(2\text{x}-1)+\log(2\text{x}-1)\frac{\text{d}}{\text{dx}}\big(\text{x}^2\big)\Big]$
[Using product rule and chain rule]
$=\frac{3}{3\text{x}+2}\Big[\text{x}^2\times\frac{1}{2\text{x}-1}\frac{\text{d}}{\text{dx}}(2\text{x}-1)+\log(2\text{x}-1)\times2\text{x}\Big]$
$=\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$
So,
$\frac{\text{d}}{\text{dx}}\big(\log(3\text{x}+2)-\text{x}^2\log(2\text{x}-1)\big) \\ =\frac{3}{3\text{x}+2}-\frac{2\text{x}^2}{2\text{x}-1}-2\text{x}\log(2\text{x}-1)$

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