parheliubdr

2023-02-27

What is the derivative of $-\mathrm{csc}\left(x\right)$?

Shiloh Hinton

Cosecant is the reciprocal of the sine function, so:
$-\frac{d}{dx}\left(\frac{1}{\mathrm{sin}\left(x\right)}\right)$
You should know that the derivative of $\frac{1}{x}=-\frac{1}{{x}^{2}}$
Thus, by the chain rule: $\frac{1}{f\left(x\right)}=-\frac{1}{{f\left(x\right)}^{2}\cdot }\frac{d}{dx}f\left(x\right)$
Applying this here:
$-\left(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \frac{d}{dx}\mathrm{sin}\left(x\right)\right)$
$=-\left(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \mathrm{cos}\left(x\right)\right)$
$=\frac{\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right)}d$
This is already an answer, but it appears in most textbooks in a different form:
$\frac{1}{\mathrm{sin}\left(x\right)}\cdot \frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$
$=\mathrm{csc}\left(x\right)\cdot \mathrm{cot}\left(x\right)$

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