Question

Finding the derivative of \(\ \large f(x)=2sinx+sin^2x ?\)

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

Okay!
So what I did was correct?

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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".

Answer 11

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

Answer 12

Sorry, I meant 3

Answer 13

Where does the 3 come from?

Answer 14

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.

Answer 15

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

Answer 16

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".