Prove that d/dx (cosecx) = -cosecxcotx
Since,
$\displaystyle \csc x=\frac{1}{\sin x}$
Then, when you take its derivative apply quotient rule,
$\displaystyle (\csc x)'=\frac{(1)'\sin x+1(\sin x)'}{\sin^2 x}$
Then,
$\displaystyle (\csc x)'=\frac{0\cdot \sin x-1\cdot \cos x}{\sin^2 x}=\frac{-\cos x}{\sin ^2x}$=$\displaystyle \frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot x\sin x$
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For reference,
If $\displaystyle f(x)$ is a diffrenciable function at $\displaystyle x=c$ and $\displaystyle g(x)$ is a diffrenciable function at $\displaystyle x=c$ such as, $\displaystyle g(c)\not = 0$, then $\displaystyle f(x)/g(x)$ is diffrenciable at $\displaystyle x=c$ and its derivative at $\displaystyle x=c$is,
$\displaystyle \frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}$