# How do you find the derivative of x/(e^(2x))?

Mar 18, 2018

$\frac{1 - 2 x}{{e}^{2 x}}$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{Given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = {e}^{2 x} \Rightarrow h ' \left(x\right) = {e}^{2 x} \times \frac{d}{\mathrm{dx}} \left(2 x\right) = 2 {e}^{2 x}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{x}{{e}^{2 x}}\right)$

$= \frac{{e}^{2 x} - 2 x {e}^{2 x}}{{e}^{2 x}} ^ 2$

$= \frac{{e}^{2 x} \left(1 - 2 x\right)}{{e}^{2 x}} ^ 2 = \frac{1 - 2 x}{{e}^{2 x}}$

Mar 18, 2018

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

#### Explanation:

we can arrange this function so that we can use the product rule

$y = \frac{x}{{e}^{2 x}}$

$\implies y = x {e}^{- 2 x}$

the product rule

$\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

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

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