# How do you find the derivative for y= ((e^x)/(x^2))?

Jul 30, 2015

You use the quotient rule.

#### Explanation:

Notice that your function can be written as the quotient of two other functions, let's say $f \left(x\right)$ and $g \left(x\right)$, which means that you can differentiate it by using the quotient rule.

color(blue)(y^' = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2, where $g \left(x\right) \ne 0$

In your case, you have $y = f \frac{x}{g} \left(x\right)$, where

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

This means that your derivative will look like this

${y}^{'} = \frac{\frac{d}{\mathrm{dx}} \left({e}^{x}\right) \cdot {x}^{2} - {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)}{{x}^{2}} ^ 2$

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

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$, so you get

${f}^{'} = \frac{{e}^{x} \cdot {x}^{2} - {e}^{\cdot} \left(2 x\right)}{x} ^ 4 = \frac{x \cdot {e}^{x} \cdot \left(x - 2\right)}{x} ^ 4$

You can simplify this to be

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot {e}^{x} \left(x - 2\right)}{x} ^ \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \textcolor{b l u e}{3}\right) = \textcolor{g r e e n}{\frac{{e}^{x} \left(x - 2\right)}{x} ^ 3}$