# What is the derivative of f(x) = cot(x)?

Oct 28, 2015

Derivative of $\cot \left(x\right)$ is equal to $- {\csc}^{2} \left(x\right)$.

#### Explanation:

We know that $\cot \left(x\right) = \frac{1}{\tan} \left(x\right)$ so $f ' \left(x\right) = \frac{1}{\tan} \left(x\right) \mathrm{dx}$

We can use the quotient rule to solve for the derivative. The quotient rule states:

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

in our case,

$g \left(x\right) = 1$
$h \left(x\right) = \tan \left(x\right)$
$g ' \left(x\right) = 0$
$h ' \left(x\right) = {\sec}^{2} \left(x\right)$

Let's plug these values back into the quotient rule:

$\frac{0 \cdot \tan \left(x\right) - 1 \cdot {\sec}^{2} \left(x\right)}{\tan} ^ 2 \left(x\right) = - {\sec}^{2} \frac{x}{\tan} ^ 2 \left(x\right) = - {\csc}^{2} \left(x\right)$