# Re: Derivatives of Trig Functions

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Posted by T.Gracken on October 21, 2002 at 16:25:07:

In Reply to: Derivatives of Trig Functions posted by MFrizza on October 21, 2002 at 13:35:09:

: Would someone please take a look at this and tell me where I've gotten off track...?

: Find the points on a curve y = (cos x)/(2 + sin x) at which the tangent is horizontal.

: (using quotient formula), y' = [(2 + sin x)(-sin x) - (cos x)(cos x)] /(2 + sin x)^2

first step looks good, but be careful when simplifying numerator...

y' = [-2sin(x) - sin^2(x) - cos^2(x)]/[4 + 4sin(x) + sin^2(x)]

so y' = -[sin(x)+1]/[2+sin(x)]2 ...leave denominator factored (my preference). I just factored out the -1 from numerator and used the pythagorean identity to rewrite the sin2(x) + cos2(x) as 1.

now, possible points on curve at which tangent is horizontal occur when y' = 0, so the "possible" points where tangent is horizontal occur for values of x such that sin(x)+1=0 ...fraction is equal to zero when numerator is equal to zero (and denominator is not).

this should lead to x = 3pi/2 + 2pi*n (n an integer)

hope that helps.

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