# How do you differentiate f(x)=sqrt((1-xe^(2x))^3 using the chain rule.?

Oct 11, 2016

$- \frac{3}{2} \sqrt{1 - x {e}^{2 x}} {e}^{2 x} \left(1 + 2 x\right)$

#### Explanation:

$f \left(x\right) = \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}$

$\left[1\right] \text{ } {D}_{x} \left[\sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}\right]$

The derivative of $\sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$

$\left[2\right] \text{ } = \frac{1}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot {D}_{x} \left[{\left(1 - x {e}^{2 x}\right)}^{3}\right]$

Use power rule

$\left[3\right] \text{ } = \frac{1}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot 3 {\left(1 - x {e}^{2 x}\right)}^{2} \cdot {D}_{x} \left[1 - x {e}^{2 x}\right]$

$\left[4\right] \text{ } = \frac{3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{D}_{x} \left[1\right] - {D}_{x} \left[x {e}^{2 x}\right]\right]$

The derivative of a constant is 0

$\left[5\right] \text{ } = \frac{3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[- {D}_{x} \left[x {e}^{2 x}\right]\right]$

$\left[6\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{D}_{x} \left[x {e}^{2 x}\right]\right]$

Use product rule

$\left[7\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{D}_{x} \left[x\right] \cdot {e}^{2 x} + x \cdot {D}_{x} \left[{e}^{2 x}\right]\right]$

The derivative of $x$ is $1$

$\left[8\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{e}^{2 x} + x \cdot {D}_{x} \left[{e}^{2 x}\right]\right]$

You can think of ${e}^{2 x}$ as ${\left({e}^{x}\right)}^{2}$

$\left[9\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{e}^{2 x} + x \cdot {D}_{x} \left[{\left({e}^{x}\right)}^{2}\right]\right]$

Use power rule

$\left[10\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left[{e}^{2 x} + x \cdot 2 {e}^{x} \cdot {D}_{x} \left[\left({e}^{x}\right)\right]\right]$

The derivative of ${e}^{x}$ is also ${e}^{x}$

$\left[11\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left({e}^{2 x} + x \cdot 2 {e}^{x} \cdot {e}^{x}\right)$

$\left[12\right] \text{ } = \textcolor{red}{\frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 \sqrt{{\left(1 - x {e}^{2 x}\right)}^{3}}} \cdot \left({e}^{2 x} + 2 x {e}^{2 x}\right)}$

You could leave it there or you can simplify it further if you are required to.

$\left[13\right] \text{ } = \frac{- 3 {\left(1 - x {e}^{2 x}\right)}^{2}}{2 {\left(1 - x {e}^{2 x}\right)}^{\frac{3}{2}}} \cdot \left({e}^{2 x} + 2 x {e}^{2 x}\right)$

$\left[14\right] \text{ } = - \frac{3}{2} {\left(1 - x {e}^{2 x}\right)}^{\frac{1}{2}} \cdot \left({e}^{2 x} + 2 x {e}^{2 x}\right)$

$\left[15\right] \text{ } = \textcolor{red}{- \frac{3}{2} \sqrt{1 - x {e}^{2 x}} {e}^{2 x} \left(1 + 2 x\right)}$