Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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?
Sorry, I meant 3
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".

Not the answer you are looking for?

Search for more explanations.

Ask your own question