# How do you find the derivative of f(x)= 1/(x-7) ?

Jun 21, 2016

$- \frac{1}{x - 7} ^ 2$ or -1/(x^2-14x+49

#### Explanation:

The quotient rule : $\frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{f ' \left(x\right) \cdot g \left(x\right) - g ' \left(x\right) \cdot f \left(x\right)}{g \left(x\right)} ^ 2$ where $f \left(x\right)$ is the numerator and $g \left(x\right)$ is the denominator

so plugging in our values:

$\frac{\left[\frac{d}{\mathrm{dx}} 1 \cdot \left(x - 7\right)\right] - \left[\frac{d}{\mathrm{dx}} \left(x - 7\right) \cdot 1\right]}{x - 7} ^ 2$ $\to$ $\frac{\left[0 \cdot \left(x - 7\right)\right] - \left[1 \cdot 1\right]}{x - 7} ^ 2$

which simplifies to:
$- \frac{1}{x - 7} ^ 2$

also quick tip: the first derivative of $\frac{1}{n}$ will always be $- \frac{1}{n} ^ 2$ as long as $n$ is always ${n}^{1}$ for basic functions like $\frac{1}{x - 7}$