# How do you find the derivative of (cot^(2)x)?

Sep 1, 2015

$- 2 {\text{cosec}}^{2} x . \cot x$

#### Explanation:

$f \left(x\right) = {\cot}^{2} x$

$d \frac{f \left(x\right)}{\mathrm{dx}} = 2 \cot x . \left(- {\text{cosec}}^{2} x\right) .1$

$= - 2 {\text{cosec}}^{2} x . \cot x$

first you get the derivative of the indice, then the trig function, lastly the x.
it's easy if you remember it that way ;)

May 29, 2017

$- \frac{2 \cot x}{\sin} ^ 2 x$

#### Explanation:

I like the previous answer, but I think it's simpler to use the product rule. Since ${\cot}^{2} x = \cot x \cdot \cot x$, $\frac{d}{\mathrm{dx}} \left({\cot}^{2} x\right) = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$, where $f \left(x\right) = g \left(x\right) = \cot x$. This evaluates to $\cot x \times \left(- \frac{1}{\sin} ^ 2 x\right) + \left(- \frac{1}{\sin} ^ 2 x\right) \times \cot x = 2 \left(- \frac{1}{\sin} ^ 2 x\right) \cot x$, $- \frac{2 \cot x}{\sin} ^ 2 x$. This is equivalent to the previous answer, since $\frac{1}{\sin} ^ 2 x = {\text{cosec}}^{2} x$, but I think that that answer is cleaner, don't you?