# How do you integrate y=(5e^-x)/(x+e^(-2x)) using the quotient rule?

Nov 19, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 5 {e}^{-} x \left(x + 1\right) + 5 {e}^{- 3 x}}{x + {e}^{- 2 x}} ^ 2$

#### Explanation:

I think you meant "how do you differentiate using the quotient rule".

If $y = f \frac{x}{g} \left(x\right)$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

Our $f \left(x\right) = 5 {e}^{-} x$ and $g \left(x\right) = x + {e}^{- 2 x}$. Then:

$f ' \left(x\right) = - 5 {e}^{-} x$ and $g ' \left(x\right) = 1 - 2 {e}^{- 2 x}$

Now let's plug them in:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 5 x {e}^{-} x - 5 {e}^{- 3 x} - 5 {e}^{-} x + 10 {e}^{- 3 x}}{x + {e}^{- 2 x}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 5 {e}^{-} x \left(x + 1\right) + 5 {e}^{- 3 x}}{x + {e}^{- 2 x}} ^ 2$