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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives3 Marks
Question
Find the derivative of $\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$
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Answer

Here $f(x) = \frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{2}{{x + 1}} - \frac{{{x^2}}}{{3x - 1}}} \right]$$= \frac{d}{{dx}}\left( {\frac{2}{{x + 1}}} \right) - \frac{d}{{dx}}\left( {\frac{{{x^2}}}{{3x - 1}}} \right)$
$= \frac{{(x + 1)\frac{d}{{dx}}(2) - 2\frac{d}{{dx}}(x + 1)}}{{{{(x + 1)}^2}}}$$- \frac{{(3x - 1)\frac{d}{{dx}}({x^2}) - {x^2}\frac{d}{{dx}}(3x - 1)}}{{{{(3x - 1)}^2}}}$
$ = \frac{{(x + 1) \times 0 - 2 \times 1}}{{{{(x + 1)}^2}}} - \frac{{(3x - 1)(2x) - {x^2 } \times 3}}{{{{(3x - 1)}^2}}}$
$= \frac{{ - 2}}{{{{(x + 1)}^2}}} - \frac{{6{x^2} - 2x - 3{x^2}}}{{{{(3x - 1)}^2}}}$$ = \frac{{ - 2}}{{{{(x + 1)}^2}}} - \frac{{3{x^2} - 2x}}{{{{(3x - 1)}^2}}}$

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