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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of $\large\frac{2}{x+1}$$ - \large\frac{x^2}{3x-1}$

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Let $f(x) =\large\frac{2}{x+1}$$ - \large\frac{x^2}{3x-1}$
$f'(x) =\large\frac{d}{dx}$$\bigg(\large\frac{2}{x+1}\bigg)$$ -\large\frac{d}{dx}$$ \bigg( \large\frac{x^2}{3x-1}\bigg)$
By quotient rule,
$ f'(x) = \bigg[ \large\frac{(x+1) \large\frac{d}{dx}(2)-2\large\frac{d}{dx}(x+1)}{(x+1)^2}\bigg]$$ \bigg[ \large\frac{(3x-1) \large\frac{d}{dx}(x^2)-x^2\large\frac{d}{dx}(3x-1)}{(3x-1)^2}\bigg]$
$ = \bigg[ \large\frac{(x+1)(0)-2(1)}{(x+1)^2} \bigg]$$- \bigg[ \large\frac{(3x-1)(2x)-(x^2)(3)}{(3x-1)^2} \bigg]$
$ = \large\frac{-2}{(x+1)^2}$$-\bigg[ \large\frac{6x^2-2x-3x^2}{(3x-1)^2} \bigg]$
$ = \large\frac{-2}{(x+1)^2}$$-\bigg[ \large\frac{3x^2-2x^2}{(3x-1)^2} \bigg]$
$ = \large\frac{-2}{(x+1)^2}$$- \large\frac{x(3x-2)}{(3x-1)^2} $
answered Apr 9, 2014 by thanvigandhi_1
 

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