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

May 26, 2016

$\setminus \frac{{e}^{x}}{{x}^{2}} - \setminus \frac{2 {e}^{x}}{{x}^{3}}$

Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left({e}^{x} \setminus \frac{1}{{x}^{2}}\right)$

Applying product rule,${\left(f \cdot g\right)}^{'} = {f}^{'} \cdot g + f \cdot {g}^{'}$

$f = {e}^{x} , g = \setminus \frac{1}{{x}^{2}}$

$= \setminus \frac{d}{\mathrm{dx}} \setminus \left({e}^{x} \setminus\right) \setminus \frac{1}{{x}^{2}} + \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \frac{1}{{x}^{2}} \setminus\right) {e}^{x}$

We know,
$\setminus \frac{d}{\mathrm{dx}} \setminus \left({e}^{x} \setminus\right) = {e}^{x}$ and

$\setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \frac{1}{{x}^{2}} \setminus\right) = - \setminus \frac{2}{{x}^{3}}$

So,
$= {e}^{x} \setminus \frac{1}{{x}^{2}} + \setminus \left(- \setminus \frac{2}{{x}^{3}} \setminus\right) {e}^{x}$

Finally, simplifying it,
$= \setminus \frac{{e}^{x}}{{x}^{2}} - \setminus \frac{2 {e}^{x}}{{x}^{3}}$