parheliubdr

2023-02-27

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

Shiloh Hinton

Beginner2023-02-28Added 9 answers

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:

$-(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \frac{d}{dx}\mathrm{sin}\left(x\right))$

$=-(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \mathrm{cos}\left(x\right))$

$=\frac{\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right)}d$

This is already an answer, but it appears in most textbooks in a different form:

$\frac{1}{\mathrm{sin}\left(x\right)}\cdot \frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$

$=\mathrm{csc}\left(x\right)\cdot \mathrm{cot}\left(x\right)$

$-\frac{d}{dx}\left(\frac{1}{\mathrm{sin}\left(x\right)}\right)$

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:

$-(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \frac{d}{dx}\mathrm{sin}\left(x\right))$

$=-(-\frac{1}{{\mathrm{sin}}^{2}\left(x\right)}\cdot \mathrm{cos}\left(x\right))$

$=\frac{\mathrm{cos}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right)}d$

This is already an answer, but it appears in most textbooks in a different form:

$\frac{1}{\mathrm{sin}\left(x\right)}\cdot \frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$

$=\mathrm{csc}\left(x\right)\cdot \mathrm{cot}\left(x\right)$