Maximizing a Trigonometric Function: How to Use the First Derivative Test

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To maximize the trigonometric function f(x) = sin(x) / (1 + cos^2(x)), the first derivative test is essential. The first step involves differentiating the function using the quotient rule and chain rule, leading to f'(x) = [cos(x)(1 + cos^2(x)) + 2sin^2(x)cos(x)] / (1 + cos^2(x))^2. Setting the numerator equal to zero will help identify critical points for potential maxima or minima. It’s noted that one solution may involve the condition 1 + cos^2(x) = 0, which indicates an asymptote. Understanding these steps is crucial for effectively applying the first derivative test in trigonometric functions.
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Hi I need help using the First derivative test on this problem: f(x)= sinx divided by 1 + cos^2x . any help would be awesome.:confused:
 
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What's the first derivative test?
I've heard of the second derivative test... but never the first..
Do you mean finding the minimum/maximum?
 
yes That is what I eventually need, I can usually get the answer after I find f '(x) but the 1+ cos^2x is the part I can't figure out can you help?
 
OK,
First, this should be in the homework section, but nobody likes a slut, so just disregard this sentence. :D
f(x) = [sin x]/(1 + cos^2 x)
Differentiate using the quotient rule and the chain rule.. (for cos^2 x, which would mean (cos x)^2 which would give you the derivative 2(-sin x)(cos x) = -2sin x cos x)
Thus, you have:
f'(x) = (cos x)(1 + cos^2 x) + 2 sin^2 x cos x all over (1 + cos^2 x)^2...
Set that equal to 0 and ignore the denominator for the moment... and see what solutions you get...
Chances are one of them might just be when 1 + cos^2 x = 0 (I don't know anything about trig functions more than the basics without a calculator, so don't approach me and prove me wrong, I'm trying to help :P) and... yeah... I guess that'll be an asymptote...
Hope this helped some.
 
Thank you for your help that made more since than what I got from class :P
 

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