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Study23
Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)
derivative of the sine is the cosine. For the second term, use the chain rule and get derivative of u^2 where u = sin x. Then multiply by derivative of u.
\(\ \text{Here's what I've done so far: } \) \(\ =sinx\times(2+sinx) \) \(\ =cosx\times(2+sinx)+sinx(cosx) \) \(\ =2cosx+sinxcosx+sinxcosx\) \(\ \text{Now what?} \)
= 2cosx + 2sinxcosx = (2cosx)(1 + sinx)
Okay! So what I did was correct?
So, the next part of this problem is to find all points at which the tangent line is horizontal. How would I find where x is zero for this derivative?
I couldn't follow your steps 2 and 3, but you were getting the right answer on step 3.
What I did for the initial steps was make sin^2 x = u. So, the derivative of u^2 is 2u and then you have to multiply by u'. So, the derivative of the second term is 2uu' or 2sinxcosx
So how I find the points at which this derivative is zero?
\(\ \Huge \text{Would it be: } \Huge \pm\frac{\pi}{2} + 2\pi k? \)
The fully factored form of the derivative is (2cosx)(1 + sinx) which is 0 at pi/2 + kpi, at every "straight up" or every "straight down".
The correct answer is apparently \(\ ((\frac{\pi}{2}+2\pi k),3) .\) Where does that 2 come from?
Where does the 3 come from?
Not 2pik, it is pik. Straight up AND straight down. cos 3pi/2 is also 0, not just pi/2. Go back and look at my answer.
Sorry, that \(\ 2\pi k \) was a typo. Im not getting from where that 3 came from
At pi/2 + 2pik you get 2 for 2sinx and 1 for sin^2 x. Added you get 3. Now for 3pi/2 + 2pik you get -2 for 2sinx and 1 for sin^2 x. Added you get -1. So, you still get derivative of 0 at pi/2 + pik (every straight up AND straight down), but you will get different values for f(x): 3 for "up" and -1 for "down".