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Study23
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Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)
 one year ago
 one year ago
Study23 Group Title
Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)
 one year ago
 one year ago

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tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \text{Here's what I've done so far: } \) \(\ =sinx\times(2+sinx) \) \(\ =cosx\times(2+sinx)+sinx(cosx) \) \(\ =2cosx+sinxcosx+sinxcosx\) \(\ \text{Now what?} \)
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
= 2cosx + 2sinxcosx = (2cosx)(1 + sinx)
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay! So what I did was correct?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
I couldn't follow your steps 2 and 3, but you were getting the right answer on step 3.
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So how I find the points at which this derivative is zero?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \Huge \text{Would it be: } \Huge \pm\frac{\pi}{2} + 2\pi k? \)
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
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".
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
The correct answer is apparently \(\ ((\frac{\pi}{2}+2\pi k),3) .\) Where does that 2 come from?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, I meant 3
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Where does the 3 come from?
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, that \(\ 2\pi k \) was a typo. Im not getting from where that 3 came from
 one year ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
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".
 one year ago
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