Derivative of $-\csc(2x)$

Question

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

Do you mean $-\csc^2(x)$ or $-\csc(2x)$? As it stands, this is very ambiguous. — alexwlchan, Sep 28 '14 at 18:00
@AlexR: If you look in the edit history, the “edit approved” link above the editor name shows who approved the edit (in this case, the OP). — alexwlchan, Sep 28 '14 at 18:05
@Hello You may want to take a look at this help page for a few tips on MathJax. — AlexR, Sep 28 '14 at 18:08

score 1 · Answer 1 · answered Jan 12 '21 at 03:20

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.

score 1 · Answer 2 · edited Sep 28 '14 at 18:04

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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.

edited Sep 28 '14 at 18:04

AlexR

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answered Sep 28 '14 at 18:01

FreeMind

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Wouldn't I need to take the derivative again of the 2x due to the chain rule? – Hello Sep 28 '14 at 18:01
You only take the derivative of $2x$ once, since that's the inner function of $-\csc(2x)$. – amWhy Sep 28 '14 at 18:02
@Hello No, you take the derivative of $2x$ which is $2$ and multiply it by the rest. – FreeMind Sep 28 '14 at 18:02
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@FreeMind You may want to take a look at this help page for a few tips on MathJax. – AlexR Sep 28 '14 at 18:04
@amWhy I edited my answer before you posting it. – FreeMind Sep 28 '14 at 18:05

score 0 · Accepted Answer · answered Sep 28 '14 at 18:00

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$$\frac{d}{dx}(-\csc (2x))= -\frac{d}{dx}(\csc (2x)) = -\big(-\csc(2x)\cot(2x)\big)\cdot \frac{d}{dx}(2x)$$

$$ = 2\csc (2x)\cot(2x)$$

answered Sep 28 '14 at 18:00

amWhy

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Derivative of $-\csc(2x)$

3 Answers3