Find the derivative of 2/x+1 - x²/3x-1 with respect to x.                  

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$$\begin{gathered} \mathrm{y}=\frac{2}{\mathrm{x}+1}-\frac{{\mathrm{x}}^2}{3\mathrm{x}-1} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2.\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}+1}\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\mathrm{x}}^2}{3\mathrm{x}-1}\right] \\ \mathrm{In} \mathrm{second} \mathrm{term} \mathrm{we} \mathrm{will} \mathrm{differentiate} \mathrm{with} \mathrm{the} \mathrm{help} \mathrm{of} \mathrm{quotient} \mathrm{rule}: \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2.\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}+1\right)}^{-1}-\frac{\left(3\mathrm{x}-1\right).{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left({\mathrm{x}}^2\right)-{\mathrm{x}}^2.{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(3\mathrm{x}-1\right)}{{\left(3\mathrm{x}-1\right)}^2} \\ =-2{\left(\mathrm{x}+1\right)}^{-2}-\frac{\left(3\mathrm{x}-1\right).\left(2\mathrm{x}\right)-{\mathrm{x}}^2.\left(3\right)}{{\left(3\mathrm{x}-1\right)}^2} \\ =-\frac{2}{{\left(\mathrm{x}+1\right)}^2}-\left[\frac{6{\mathrm{x}}^2-2\mathrm{x}-3{\mathrm{x}}^2}{{\left(3\mathrm{x}-1\right)}^2}\right] \\ =-\frac{2}{{\left(\mathrm{x}+1\right)}^2}-\left[\frac{3{\mathrm{x}}^2-2\mathrm{x}}{{\left(3\mathrm{x}-1\right)}^2}\right] \\ \Rightarrow \boxed{\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{{\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}}\mathbf{-}\left[\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}}{{\left(\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}}\right]} \end{gathered}$$

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