kreiranihqlz

2023-03-09

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

zarizaraje1

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

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