# What is the derivative of this function -1/(sin(x))^2?

Dec 3, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cot x {\csc}^{2} x$

#### Explanation:

This function can be written as $y = - {\left(\sin x\right)}^{-} 2$.

We let $y = - {u}^{-} 2$ and $u = \sin x$. Through the power rule and basic differentiation rules, we obtain $\frac{\mathrm{dy}}{\mathrm{du}} = - 2 \left(- 1\right) {u}^{-} 3 = \frac{2}{u} ^ 3$ and $\frac{\mathrm{du}}{\mathrm{dx}} = - \cos x$.

Recall that the chain rule for derivatives of composite functions states that $\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{u} ^ 3 \times \cos x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos x}{\sin x} ^ 3$

By the identities $\cot x = \frac{1}{\tan} x$ and $\csc x = \frac{1}{\sin} x$, we can simplify the derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cot x {\csc}^{2} x$

Hopefully this helps!