Question

Find the derivative of
$\frac{2}{x+1}-\frac{x^2}{3x-1}$

Answer 1

Let $f\left(x\right)=\frac{2}{x+1}-\frac{x^2}{3x-1}$
$f^{{}^{\prime }}\left(x\right)=\frac{d}{dx}\left(\frac{2}{x+1}\right)-\frac{d}{dx}\left(\frac{x^2}{3x-1}\right)$
By quotient rule
$f^{{}^{\prime }}\left(x\right)=\left[\frac{(x+1)\frac{d}{dx}\left(2\right)-2\frac{d}{dx}(x+1)}{{(x+1)}^2}\right]-\left[\frac{(3x-1)\frac{d}{dx}\left(x^2\right)\frac{d}{dx}(3x-1)}{{(3x-1)}^2}\right]$
$=\left[\frac{(x+1)\left(0\right)-2\left(1\right)}{{(x+1)}^2}\right]-\left[\frac{(3x-1)\left(2x\right)-\left(x^2\right)\left(3\right)}{{(3x-1)}^2}\right]$
$=\frac{-2}{{(x+1)}^2}-\left[\frac{6x^2-2x-3x^2}{{(3x-1)}^2}\right]$
$=\frac{-2}{{(x+1)}^2}-\left[\frac{3x^2-2x}{{(3x-1)}^2}\right]$
$=\frac{-2}{{(x+1)}^2}-\frac{x(3x-2)}{{(3x-1)}^2}$