# How do you find the derivative of y=e^(-2x-1) * ln(-2x-1)?

Jun 21, 2018

$- {e}^{- 2 x - 1} / \left(- 2 x - 1\right) - 2 \ln \left(- 2 x - 1\right) {e}^{- 2 x - 1}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \ln \left(- 2 x - 1\right) {e}^{- 2 x - 1}$
=$\frac{d}{\mathrm{dx}} \left[\ln \left(- 2 x - 1\right)\right] \cdot {e}^{- 2 x - 1} + \ln \left(- 2 x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{- 2 x - 1}\right)$
=$\frac{1}{- 2 x - 1} \cdot \frac{d}{\mathrm{dx}} \left(- 2 x - 1\right) \cdot {e}^{- 2 x - 1} + {e}^{- 2 x - 1} \cdot \frac{d}{\mathrm{dx}} \left[- 2 x - 1\right] \cdot \ln \left(- 2 x - 1\right)$
=$\frac{\left(- 2 \cdot \frac{d}{x} \left[x\right] + \frac{d}{\mathrm{dx}} \left[- 1\right]\right) {e}^{- 2 x - 1}}{- 2 x - 1} + \left(- 2 \cdot \frac{d}{x} \left[x\right] + \frac{d}{\mathrm{dx}} \left[- 1\right]\right) \ln \left(- 2 x - 1\right) {e}^{- 2 x - 1}$
=$\frac{\left(0 - 2 \cdot 1\right) {e}^{- 2 x - 1}}{- 2 x - 1} + \left(0 - 2 \cdot 1\right) \ln \left(- 2 x - 1\right) {e}^{- 2 x - 1}$
=$- {e}^{- 2 x - 1} / \left(- 2 x - 1\right) - 2 \ln \left(- 2 x - 1\right) {e}^{- 2 x - 1}$