LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
Find the derivative of. $\frac{2}{\text{x}-1}-\frac{\text{x}^2}{3\text{x}-1}$
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Answer
Let $\text{f}(\text{x})=\frac{2}{\text{x}-1}-\frac{\text{x}^2}{3\text{x}-1}$ $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{2}{\text{x}+1}\bigg)-\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^2}{3\text{x}-1}\bigg)$ By quotient rule, $\text{f}'\big(\text{x})=\Bigg[\frac{(\text{x}+1)\frac{\text{d}}{\text{dx}}(2)-2\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]-\Bigg[\frac{(3\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}^2)-\text{x}^2\frac{\text{d}}{\text{dx}}(3\text{x}-1)}{(3\text{x}+1)^2}\Bigg]$$=\bigg[\frac{(\text{x}+1)(0)-2(1)}{(\text{x}+1)^2}\bigg]-\bigg[\frac{(3\text{x}-1)(2\text{x})-(\text{x}^1)(3)}{(3\text{x}-1)^2}\bigg]$
$=\frac{-2}{(\text{x}+1)^2}-\bigg[\frac{6\text{x}^2-2\text{x}-3\text{x}^2}{(3\text{x}-1)^2}\bigg]$
$=\frac{-2}{(\text{x}+1)^2}-\bigg[\frac{3\text{x}^2-2\text{x}^2}{(3\text{x}-1)^2}\bigg]$
$=\frac{2}{(\text{x}+1)^2}-\frac{\text{x}(3\text{x-2})}{(3\text{x}-1)^2}$
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