Is the following reasoning correct? $$-\csc2x = -1[\csc(2x)\cdot\cot(2x)] = -1[\csc(2x)\cdot\cot(2x)\cdot2\cdot2]$$
I am unsure whether that is correct so far because I do not know if the derivatives of the $\csc$ and $\cot$ need to be taken again due to the chain rule
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$\begingroup$The derivative of the $\csc(x) = -\cot(x)\cdot\csc(x)$
The answer to your question is $2\cot(2x)\csc(2x)$
Simply, you have to use the chain rule.
First of all take the derivative of $\csc$ , then the $2x$ and multiply them.
Why don't you simply use sine and cosine?
The derivative can then be easily calculated using basic differentiation rules:$$ -\frac{d}{dx}\csc(2x)=\frac{d}{dx}\frac{-1}{\sin(2x)}=\frac{1}{\sin^2(2x)}\cdot \cos(2x)\cdot 2=2\cos(2x)\csc(2x) $$
"Is the following reasoning correct?" NO. You miss the derivative symbol, not to say other aspects.
$\endgroup$ $\begingroup$$$\frac{d}{dx}(-\csc (2x))= -\frac{d}{dx}(\csc (2x)) = -\big(-\csc(2x)\cot(2x)\big)\cdot \frac{d}{dx}(2x)$$
$$ = 2\csc (2x)\cot(2x)$$
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