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

Jan 13, 2016

$\frac{{e}^{x} \left(x - 2\right)}{{x}^{3}}$ with the quotient rule.

#### Explanation:

You can use the quotient rule to find the derivative.

If $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then the derivative is:

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$

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

The derivatives of $g \left(x\right)$ and $h \left(x\right)$ are

$g ' \left(x\right) = {e}^{x}$ and $h ' \left(x\right) = 2 x$.

Thus, according to the formula, you derivative is:

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

... cancel $x$ both in the numerator and the denominator...

$= \frac{\cancel{\textcolor{b l u e}{x}} {e}^{x} \left(x - 2\right)}{{x}^{3} \cdot \cancel{\textcolor{b l u e}{x}}} = \frac{{e}^{x} \left(x - 2\right)}{{x}^{3}}$