# How do you differentiate g(x) = (1/x^3)*sqrt(1-e^(2x)) using the product rule?

Feb 13, 2018

$\setminus$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g ' \left(x\right) \setminus = \setminus \frac{3 + \left(x - 3\right) {e}^{2 x}}{{x}^{4} \setminus \sqrt{1 - {e}^{2 x}}} \setminus \quad .$

#### Explanation:

$\setminus$

$\text{We are given:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g \left(x\right) \setminus = \setminus \left(\frac{1}{x} ^ 3\right) \setminus \cdot \setminus \sqrt{1 - {e}^{2 x}} .$

$\text{We can rewrite" \ g(x) \ "to prepare it for differentiation:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g \left(x\right) \setminus = \setminus {x}^{- 3} \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}} .$

$\text{Using the Product Rule:}$

$\setminus q \quad \setminus q \quad g ' \left(x\right) \setminus = \setminus {x}^{- 3} \setminus \cdot \left[{\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}}\right] ' + \left[{x}^{- 3}\right] ' \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}} .$

$\text{Using the Chain Rule twice on the first differentiated quantity:}$

$g ' \left(x\right) \setminus =$

$\setminus \setminus \setminus {x}^{- 3} \setminus \cdot \left[\frac{1}{2} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \left(1 - {e}^{2 x}\right) '\right] + \left[- 3 {x}^{- 4}\right] \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}}$

$\setminus q \quad \setminus q \quad \setminus =$

$\setminus \setminus \setminus {x}^{- 3} \setminus \cdot \left[\frac{1}{2} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \left(0 - 2 {e}^{2 x}\right)\right] + \left(- 3 {x}^{- 4}\right) \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}} .$

$\text{Simplify:}$

$\setminus q \quad \setminus q \quad \setminus =$

$\setminus \setminus \setminus {x}^{- 3} \setminus \cdot \left[\frac{1}{2} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \left(- 2 {e}^{2 x}\right)\right] + \left(- 3 {x}^{- 4}\right) \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}}$

$\setminus q \quad \setminus q \quad \setminus =$

$\setminus \setminus \setminus {x}^{- 3} \setminus \cdot \left[- {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \left({e}^{2 x}\right)\right] + \left(- 3 {x}^{- 4}\right) \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}} .$

$\text{Continue simplification by pulling out lowest powers of same}$
$\text{quantities:}$

$\setminus q \quad \setminus q \quad \setminus =$

$\setminus \setminus \setminus - {x}^{- 4} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \setminus \cdot \left(- {x}^{1} \left({e}^{2 x}\right) + \left(- 3\right) \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{1}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus \setminus \setminus {x}^{- 4} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \setminus \cdot \left[\left(x {e}^{2 x}\right) + 3 \setminus \cdot \left(1 - {e}^{2 x}\right)\right]$

$\setminus q \quad \setminus q \quad \setminus = \setminus \setminus \setminus {x}^{- 4} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \setminus \cdot \left[x {e}^{2 x} + 3 - 3 {e}^{2 x}\right]$

$\setminus q \quad \setminus q \quad \setminus = \setminus \setminus \setminus {x}^{- 4} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \setminus \cdot \left[3 + x {e}^{2 x} - 3 {e}^{2 x}\right]$

$\setminus q \quad \setminus q \quad \setminus = \setminus \setminus \setminus {x}^{- 4} {\left(1 - {e}^{2 x}\right)}^{- \frac{1}{2}} \setminus \cdot \left[3 + \left(x - 3\right) {e}^{2 x}\right] .$

$\text{Remove negative exponents, write one quantity as a square root:}$

$\setminus q \quad \setminus q \quad \setminus = \setminus \setminus \setminus \frac{3 + \left(x - 3\right) {e}^{2 x}}{{x}^{4} \setminus \sqrt{1 - {e}^{2 x}}} \setminus \quad .$

$\text{This is our answer.}$

$\setminus$

$\text{Summarizing:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g \left(x\right) \setminus = \setminus {x}^{- 3} \setminus \cdot {\left(1 - {e}^{2 x}\right)}^{\frac{1}{2}} .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad g ' \left(x\right) \setminus = \setminus \frac{3 + \left(x - 3\right) {e}^{2 x}}{{x}^{4} \setminus \sqrt{1 - {e}^{2 x}}} \setminus \quad .$