(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the derivative of [tex]\frac{sin x}{1 + cos x}[/tex]

2. Relevant equations

Quotient rule [tex]\frac{gf' - fg'}{g^{2}}[/tex]

3. The attempt at a solution

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{(1 + cos x)(\frac{d}{dx}(sin x)) - sin x(\frac{d}{dx}(1 + cos x)}{(1 + cos x)^{2}}[/tex]

simplify the derivative so far:

= [tex]\frac{(1 + cos x)(cos x) - (sin x)(-sin x)}{(1 + cos x)^{2}}[/tex]

simplify further:

= [tex]\frac{cos x + cos^{2}x + sin^{2}x}{(1 + cos x)^{2}}[/tex]

Use angle identity [tex]sin^{2}x[/tex] + [tex]cos^{2}x = 1[/tex]to simplify even further:

= [tex]\frac{cos x + 1}{(1 + cos)^{2}}[/tex]

cancel out the common 1 + cos x

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{1 + cos x}[/tex]

I was quite confident in my answer, but I was a little teeny bit hesitant, so I used my graphing calculator to double check. When I did so, I found out that I was wrong, the derivative that I calculated (above) is not the actual derivative of the question.

At which step did I go wrong?

Thanks so much in advance everyone!

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# Homework Help: Differentiation of a trig function using quotient rule

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