What is the derivative of (tan^-1 (x+2)/(1+2x))?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left(1 + 2 x\right) \left({x}^{2} + 4 x + 5\right)} - \frac{2}{{\left(1 + 2 x\right)}^{2}} \setminus {\tan}^{- 1} \left(x + 2\right)$

Explanation:

Given function:

$y = \setminus \frac{\setminus {\tan}^{- 1} \left(x + 2\right)}{1 + 2 x}$

differentiating w.r.t. $x$ using product rule as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{\setminus {\tan}^{- 1} \left(x + 2\right)}{1 + 2 x}\right)$

$= \setminus \frac{\left(1 + 2 x\right) \frac{d}{\mathrm{dx}} \setminus {\tan}^{- 1} \left(x + 2\right) - \setminus {\tan}^{- 1} \left(x + 2\right) \frac{d}{\mathrm{dx}} \left(1 + 2 x\right)}{{\left(1 + 2 x\right)}^{2}}$

$= \setminus \frac{\left(1 + 2 x\right) \setminus \frac{1}{1 + {\left(x + 2\right)}^{2}} - \setminus {\tan}^{- 1} \left(x + 2\right) \left(2\right)}{{\left(1 + 2 x\right)}^{2}}$

$= \frac{1}{\left(1 + 2 x\right) \left({x}^{2} + 4 x + 5\right)} - \frac{2}{{\left(1 + 2 x\right)}^{2}} \setminus {\tan}^{- 1} \left(x + 2\right)$