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

DLS Group Title

Find dy/dx !

  • one year ago
  • one year ago

  • This Question is Closed
  1. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]

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

    I want to build my logic and way of attacking problems !

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

    \[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]

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

    errr implicit derivative... only with variables? D:

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

    no actual points? xD its going to get messy

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

    haha,i know..and no sorry :D

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

    alright i'll try it out, I'm taking cal 1, good review for my test. Hum first, the whole thing is an addition of 2 big terms, so find the derivative of bot terms. The first one is a chain rule : -sin s(ax ^2+bx+c) * (2ax + b)

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

    yes!! i did that

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

    what about the sin^3 part :S

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

    secondly comes the other term. Because its a root we'll simplify it to: sin ^3 (ax ^2+bx+c)^(1/2) Again the chain rule So we get cos^3(ax ^2+bx+c)^(1/2) * (ax ^2+bx+c)^(1/2)' Therefore we have ANOTHER chain rule: 1/2 (ax ^2+bx+c)^-(1/2) * (2ax+b)

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

    so you have what I think of a chainception

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

    Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here

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

    Isnt this what we have atm? \[\frac{dy}{dx}= -\sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b\]

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

    in the end thats 2ax+b

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

    yeah thats it. Thats what you should have in my opinion :)

    • one year ago
  16. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]

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

    im doubtful about derivative of sin^3 as cos^3

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

    nope I was right :D hehehe

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

    HA! :D

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

    Sirm3d. you take the derivative of the exponent of cos too?

    • one year ago
  21. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]

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

    sugoi!

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

    ah silly me I was thinking of sin(3x) or something

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

    so u failed in ur test :P

    • one year ago
  25. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]

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

    O___O

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

    Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!

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

    okay!

    • one year ago
  29. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.

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

    why did we have a cos there?

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

    \[\large \sin^{3}x \] suppose we have this

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

    wont it be onlly 3cos^2x

    • one year ago
  33. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    3(sin x)^2 * cos x by chain rule

    • one year ago
  34. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    you should write it as (sin x)^3, then apply the chain rule

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

    okay..!

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

    thanks!

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

    i get to learn a lot from every ques! :D

    • one year ago
  38. sirm3d Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1354535975146:dw|

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

    yes i got it!

    • 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.