Differentiation of a trig function using quotient rule

stripes

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!

radou

Hm, I don't see where the mistake is. If there is one at all. Perhaps you messed up your input in the graphics calculator?

tiny-tim

Hi stripes!

Looks ok to me.

As a double-check, using standard trigonometric identities …

sinx/(1 + cosx) = tan(x/2), so dy/dx = 1/2 sec²(x/2) = 1/(1 + cosx).

What answer did your calculator give?

stripes

basically, on my graphing calculator i put y1 as my original function, then y2 as the derivative that I found. I then used the calculator to find dy/dx of y1 for me at various points (maybe 1, 2, and 3), then I used the calculator to find the values of x = 1, 2, and 3 for y2 and the respective values should be the same...but they weren't!

I'm almost certain i was error free inputting the functions...I did it over and over and over!

Mark44

Might be a dumb question, but was your calculator in radian mode?