The derivative of $\csc x$ is $(-\csc x)(\cot x) =(\frac{-1}{\sin x} ) ( \frac{\cos x}{\sin x} ) $
When I take the derivative of $ x \csc(2x-1) $, I get $\frac{x}{\sin(2x-1)}= \frac{\sin(2x-1) - 2x\cos(2x-1)}{\sin^2(2x-1)} $
How does my answer above link to the formula that I written at the top?
In the numerator , there is $\sin$ when the formula states it will only have $\cos$
Sorry I'm trying to simplify and understand $\csc$
$\endgroup$ 33 Answers
$\begingroup$Note that if $F'(x) = f(x)$, then $\frac{d}{dx} F(ax+b) = af(ax+b)$, by the chain rule. So $\frac{d}{dx}\csc(2x-1) = -2\csc(2x-1)\cot(2x-1)$. Therefore, using the product rule, $$\frac{d}{dx} x\csc(2x-1) = \csc(2x-1)\left(\frac{d}{dx} x\right)+x\left(\frac{d}{dx} \csc(2x-1) \right)$$ $$ = \csc(2x-1)-2x\csc(2x-1)\cot(2x-1)$$ $$= \frac{1-2x\cot(2x-1)}{\sin(2x-1)} = \frac{\sin(2x-1)-2x\cos(2x-1)}{\sin^2(2x-1)}$$ The last equality comes from multiplying the top and bottom by $\sin(2x-1)$.
$\endgroup$ $\begingroup$$$\begin{align}\dfrac d{dx}(\csc x)&=\dfrac d{dx}\left(\dfrac1{\sin x}\right)\\&=\dfrac d{dx}(\sin x)^{-1}\\&=-(\sin x)^{-2}\dfrac d{dx}(\sin x)\\&=-\dfrac{\cos x}{\sin^2 x}\\&=-\dfrac 1{\sin x}\cdot\dfrac{\cos x}{\sin x}\\&=-\csc x\cot x\end{align}$$
Now let's use this result, to find the derivative of your function
$$\begin{align}\dfrac d{dx}(x\csc(2x-1))&=x\dfrac d{dx}(\csc(2x-1))+\csc(2x-1)\dfrac d{dx}(x)\\&=-x\csc(2x-1)\cot(2x-1)\dfrac d{dx}(2x-1)+\csc(2x-1)\\&=-2x\csc(2x-1)\cot(2x-1)+\csc(2x-1)\\&=(1-2x\cot(2x-1))\csc(2x-1)\end{align}$$
If you change the ratios to $\sin(2x-1)$ and $\cos(2x-1)$ and simplify, or multiply top and bottom by $\sin(2x-1)$, then it will match up with the result that you have.
See if this makes things a little clear
$\endgroup$ $\begingroup$Just for fun, since your title is "Derivative̲s̲ of $\csc x$", we can note that $\forall n \in \mathbb{Z}_{\ge 0}$ $$\frac{d^n}{dx^n}\csc(x)=i^n \csc (x) \sum _{j=0}^n \sum _{k=0}^j (-1)^k k! 2^{j-k} \binom{n}{j} \left\{ {j \atop k}\right\} (1-i \cot (x))^k$$ From here, the solution is "obvious" ;)
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