# 9(vi)   Find the derivative of $\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

P Pankaj Sanodiya

Given

$f(x)=\frac{2}{x+1}- \frac{x^2 }{3 x-1}$

As we know the quotient rule of derivative:

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$

and the property

$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$

So applying this rule, we get

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{(x+1)\frac{d(2)}{dx}-2\frac{d(x+1)}{dx}}{(x+1)^2}-\frac{(3x-1)\frac{d(x^2)}{dx}-x^2\frac{d(3x-1)}{dx}}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{(3x-1)2x-x^23}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{6x^2-2x-3x^2}{(3x-1)^2}$

$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

Hence

$f'(x)=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$

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