Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Study23
Group Title
Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)
 2 years ago
 2 years ago
Study23 Group Title
Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)
 2 years ago
 2 years ago

This Question is Closed

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.
 2 years ago

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

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

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

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

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

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

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Where does the 3 come from?
 2 years 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.
 2 years 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
 2 years 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".
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.