1. ## Critical Numbers

I need some help with these problems I have to find the critical numbers fo the problems.

First I need to differentiate the problem and then I need to set it equal to 0 and solve for the variable.

The first one I just need help differentiating:

f(theta)=2cos(theta)+sin^2(theta)

This second one I differentiated but I don't know how to solve for theta:

g(theta)=4(theta)-tan(theta)
and
g'(theta)=4-sec^2(theta)

I appreciate any help.

2. ## Solution

Well, for the first one, you take the derivative of each term
Keep in mind:
The derivative of cosx = -sinx
The derivative of sinx = cosx

The first term's derivative would be 2 * -sin(theta) = -2sin(theta)
The second term's derivative involves the chain rule
It is gotten by first using the power rule, and then differentiating the trig function inside the parenthesis:
So, it is 2sin(theta)*cos(theta)

So, f'(theta) = -2sin(theta) +2sin(theta)cos(theta)

I'll leave the factoring and solving for theta to you.

3. ## Second Question

g'(theta)=4-sec^2(theta)

So, you set the right side equal to zero:
4 - sec^2(theta) = 0

Using a little algebra:
sec^2(theta) = 4

Take the square root of both sides to get:
sec(theta) = 2

Now remember, sec(theta) = 1/cos(theta)
So, 1/cos(theta) = 2

Take the inverse of both sides to get:
cos(theta) = 1/2

I'm sure you know how to do the rest. I hope this helps.
Also, for the first problem, do some review of the chain rule. It will help you understand how to get the derivatives a lot better.

4. Originally Posted by ajj86
g'(theta)=4-sec^2(theta)

So, you set the right side equal to zero:
4 - sec^2(theta) = 0

Using a little algebra:
sec^2(theta) = 4

Take the square root of both sides to get:
sec(theta) = 2

Now remember, sec(theta) = 1/cos(theta)
So, 1/cos(theta) = 2

Take the inverse of both sides to get:
cos(theta) = 1/2

I'm sure you know how to do the rest. I hope this helps.
Also, for the first problem, do some review of the chain rule. It will help you understand how to get the derivatives a lot better.
Acutally I don't know what to do next. That's where I'm stuck.

Sorry for not replying, but I haven't been on the forum since thursday or so.
The last part of the second problem involves taking the arccos of both sides to yield:

theta = arccos(1/2)

Now, we think what values of theta will cause cos(theta) to equal 1/2.

well, we know cos(Pi/3) will yield a value of 1/2. Also, since the 4th quadrant is also positive for cosine, the theta value of 5Pi/3 will yield 1/2. Now, the trick is writing these so that it covers all values that will make cos(theta) equal to 1/2.

theta = n*Pi/3 for n = 1,5,7,11,13,17,...

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### f(theta)=2 cos(theta) sin^2(theta)

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