Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

gFunc Group Title

The question for Session4 requires you to find the derivative of sin(2x). I tried using various sin/cos properties but nothing was able to cancel out the deltaX in the denominator. I tried things like sin(x+y) = sin(x)cos(y)+cos(x)sin(y), and cos(x+y) = cos(x)cos(y)-sin(x)sin(y), and sin(x)^2 = 1 - cos(x)^2. Does anyone know the property I should examine in order to help pull out the deltaX?

  • one year ago
  • one year ago

  • This Question is Closed
  1. Silent_Sorrow Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Isn't the derivative of Sin(2x) just 2Cos(2x) ?

    • one year ago
  2. gFunc Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, looking at a derivatives table I see that sin(ax) is acos(ax) but I was curious as to how to get to that conclusion. I wanted to find df/dx of sin(2x) the same way Newton did :p. I suppose I need to look back into more trig properties. Do you know of any that are particularly useful in this case. I have a feeling cos^2 + sin^2 = 1 is at play here, but I can't put my finger on it. What do you think?

    • one year ago
  3. Silent_Sorrow Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh you would just find the derivative using first principle, let f(x) = sin 2x so f(x+h) = sin(2x + h) [f(x+h) - f(x)]/h = [sin(2x + h) - sin(2x)]/h now use sin(a+b) = sin a cos b + cos a sin b so sin(2x+h) = sin 2x cos h + cos 2x sin h lim(h->0)[sin 2x cos 2h + cos 2x sin 2h - sin 2x]/h = lim(h->0) [sin 2x cos 2h - sin 2x]/h + lim(h->0) [cos 2x sin 2h]/h = sin 2x *lim(h->0) [cos 2h - 1]/h + cos 2x * lim(h->0) [sin 2h]/h = sin 2x * 0 + cos 2x * lim(h->0)[2 sin h cos h]/h = cos 2x * 2*lim(h->0)[sin h]/h * lim(h->0) cos h = 2cos 2x (1)(1) = 2cos 2x

    • one year ago
  4. SomeBloke Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    When working it out as a limit of a difference quotient, I think it's probably similar to the problem of finding the derivative of sin(x), which is dealt with in Session 7, though that still leaves a couple of tricky things for Session 8. When you've also learnt about the chain rule, you'll be able to use that in order to reduce the problem to the problem of differentiating sin(x).

    • one year ago
  5. gFunc Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahh, great feedback! I can see how d/dx of cosx at 0 is zero geometrically speaking just by looking at a cos curve but I'm lost at how lim(h->0) [cosh - 1]/h goes to 0. Is that simply because the constant (-1) goes away as a rule? I can also see how lim(h->0)sinh/h at 0 might be 1 by looking at the sine curve but I feel the need to prove it somehow. Hmm... what do you think? I recon there is a proof that the slope of sin at 0 is 1 but maybe my skills will be lacking until session 8.

    • one year ago
  6. Silent_Sorrow Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    What's taught in session 8?

    • one year ago
  7. gFunc Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    SomeBloke mentioned the chain rule in Session 8 being used to differentiate sinx.

    • one year ago
  8. SomeBloke Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Oops! I could have been clearer. The chain rule is introduced in Session 11. It lets you break differentiations down into a chain of smaller, easier, more manageable steps. Session 7 is the session in which sin(x) and cos(x) are differentiated, but the proofs rely on stuff done in Session 8. Session 8 shows the following:-\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\]\[\lim_{x \rightarrow 0}\frac{ 1-\cos x }{ x }=0.\]The proofs make use of geometry. You might also like to look up the squeeze theorem on Wikipedia. That article gives those two limits as its second example of applying the squeeze theorem, though it doesn't go into detail for them.

    • one year ago
  9. Amed Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Let z=2x by the chain rule d(sin 2x)/dx = (d(sin z)/dz)*(dz/dx) = (cos z)*(d(2x)/dx)=(cos 2x)*2=2cos(2x)

    • one year ago
  10. Sujay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Derivative of [f(x)=\sin(2x)\] Derivative of \[\sin(x)\] is \[\cos(x)\]. By the chain rule you, in simple terms, do the derivative of the outside multiplied by the derivative of the inside. The derivative of the outside is cos(2x), while the derivative of the inside is 2, multiply them together and.... the final answer is \[f'(x)=2\cos(2x) \].

    • one year ago
  11. Krishnadas Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    sin2x=2sinxcosx...now integrate by parts

    • one year ago
    • Attachments:

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.