## anonymous 3 years ago Finding the derivative of $$\ \large f(x)=2sinx+sin^2x ?$$

1. anonymous

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. anonymous

$$\ \text{Here's what I've done so far: }$$ $$\ =sinx\times(2+sinx)$$ $$\ =cosx\times(2+sinx)+sinx(cosx)$$ $$\ =2cosx+sinxcosx+sinxcosx$$ $$\ \text{Now what?}$$

3. anonymous

= 2cosx + 2sinxcosx = (2cosx)(1 + sinx)

4. anonymous

Okay! So what I did was correct?

5. anonymous

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?

6. anonymous

I couldn't follow your steps 2 and 3, but you were getting the right answer on step 3.

7. anonymous

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

8. anonymous

So how I find the points at which this derivative is zero?

9. anonymous

$$\ \Huge \text{Would it be: } \Huge \pm\frac{\pi}{2} + 2\pi k?$$

10. anonymous

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".

11. anonymous

The correct answer is apparently $$\ ((\frac{\pi}{2}+2\pi k),3) .$$ Where does that 2 come from?

12. anonymous

Sorry, I meant 3

13. anonymous

Where does the 3 come from?

14. anonymous

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.

15. anonymous

Sorry, that $$\ 2\pi k$$ was a typo. Im not getting from where that 3 came from

16. anonymous

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".