kreiranihqlz

## Answered question

2023-03-09

What is the derivative of $\frac{\mathrm{sin}\left(x\right)}{1-\mathrm{cos}\left(x\right)}$?

### Answer & Explanation

zarizaraje1

Beginner2023-03-10Added 5 answers

I would use the Quotient Rule to get:
$y=\frac{\mathrm{sin}\left(x\right)}{1-\mathrm{cos}\left(x\right)}\phantom{\rule{0ex}{0ex}}{y}^{\prime }=\frac{\mathrm{cos}\left(x\right)\left[1-\mathrm{cos}\left(x\right)\right]-\mathrm{sin}\left(x\right)\mathrm{sin}\left(x\right)}{\left[1-\mathrm{cos}\left(x\right){\right]}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)}{\left[1-\mathrm{cos}\left(x\right){\right]}^{2}}=\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(x\right)-\left[{\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)\right]}{\left[1-\mathrm{cos}\left(x\right){\right]}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\left(x\right)-1}{\left[1-\mathrm{cos}\left(x\right){\right]}^{2}}=-\frac{1-\text{⧸}cos\left(x\right)}{\left[1-\mathrm{cos}\left(x\right){\right]}^{\text{⧸}2}}\phantom{\rule{0ex}{0ex}}=-\frac{1}{1-\mathrm{cos}\left(x\right)}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?