1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Differentiation

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Differentiate the following and display in the simplest form.

    Part A:
    [tex]y=0.2x^5 - sin4x + cos4x[/tex]

    Part B:
    [tex]y=(2x^4 +3)^3
    [/tex]

    Part C:
    [tex]
    y=2x^3 sinx
    [/tex]

    Part D:
    [tex]
    y=\frac{sinx}{x^2}
    [/tex]

    Part E:
    [tex]
    y=\frac{x^3}{x^2 +1}
    [/tex]



    2. Relevant equations
    chain rule
    product rule
    quotient rule


    3. The attempt at a solution
    Part A:
    [tex]
    y=0.2x^5 - sin4x + cos4x \\
    y'=x^4-4 \ cos4x-4 \ sin4x
    [/tex]

    Part B:
    [tex]
    y=(2x^4 +3)^3 \\
    let \ u =2x^4 +3 \\
    u'=8x^3 \\
    y'(u)=3u^2 \\
    y'=3u^2 \times 8x^3 \\
    y'=3(2x^4 +3)^2 \times 8x^3 \\
    y'=24x^3(2x^4+3)^2 \\
    [/tex]

    Part C:
    [tex]
    y=2x^3 \ sinx \\
    let \ u = 2x^3 \\
    let \ v = sinx \\
    y'=vu'+uv' \\
    y'=sinx \ 6x^2 + 2x^3 \ cosx \\
    y'=2x^2(3 \ sinx + x \ cosx)
    [/tex]

    Part D:
    [tex]
    y=\frac{sinx}{x^2} \\
    y'=\frac{vu'-uv'}{v^2} \\
    y'=\frac{x^2 cosx - sinx \ 2x}{(x^2)^2} \\
    y'=\frac{x(x \ cosx - 2sinx)}{x^4} \\
    y'=x^{-3}(x \ cosx - 2sinx)
    [/tex]

    Part E:
    [tex]
    y=\frac{x^3}{x^2 +1} \\
    y'=\frac{vu'-uv'}{v^2} \\
    y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\
    y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\
    y'=\frac{x^2(3x^2 + 3 - 2x^2 )}{(x^2 +1)^2} \\
    [/tex]
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2
    A, B and C are ok . D is ok too just cancel that x you factored out with one x from the denominator to get it in simplest form. E is wrong try to rework it.
     
  4. May 14, 2013 #3
    Thanks for taking a look :), I have edited part D, is that what you meant?

    On Part E, is it my jump from line 3 to line 4 where I went wrong or was it before that?

    Thanks.
     
  5. May 14, 2013 #4

    Mark44

    Staff: Mentor

    In the next to last step in part E, simplify the numerator first by combining like terms.

    Also, it is best to post only a single problem, or two at most, in a thread.
     
  6. May 14, 2013 #5
    Here is my rework of Part E, I went around it a different way by expanding the denomator and then cancelling terms:
    Part E:
    [tex]
    y=\frac{x^3}{x^2 +1} \\
    y'=\frac{vu'-uv'}{v^2} \\
    y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\
    y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\
    y'=\frac{x^4 + 3x^2}{x^4 + 2x^2 +1} \\
    y'=\frac{3x^2}{2x^2+1}
    [/tex]

    Does this work?

    Thanks.
     
    Last edited: May 14, 2013
  7. May 14, 2013 #6
    OK no problem, will do, I just didn't want to create 3+ threads all in one go.
     
  8. May 14, 2013 #7
    No that does not work you can't just cancel out an x^4 because you haven't factored with it.

    You third to last line is correct you will get x^4+3x^2 and just factor that out and you are done.
     
  9. May 14, 2013 #8
    [tex]
    y=\frac{x^3}{x^2 +1} \\
    y'=\frac{vu'-uv'}{v^2} \\
    y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\
    y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\
    y'=\frac{x^4 + 3x^2}{(x^2 + 1)^2} \\
    y'=\frac{x^2(x^2 +3)}{(x^2 + 1)^2} \\
    [/tex]

    like that? Thanks.
     
  10. May 14, 2013 #9
    Yep that's it!
     
  11. May 14, 2013 #10
    Thank you :)
     
  12. May 14, 2013 #11
    Not a problem.
     
  13. May 22, 2013 #12
    Hi again!

    I have just been playing around with part E again and think I may have made it even simpler, can you tell me if what I have done below is OK?

    [tex]
    y=\frac{x^3}{x^2 +1} \\
    y'=\frac{vu'-uv'}{v^2} \\
    y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\
    y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\
    y'=\frac{x^4 + 3x^2}{(x^2 + 1)^2} \\
    y'=\frac{x^2(x^2 +3)}{(x^2 + 1)^2} \\
    y'=\frac{x^2(x^2 +3)}{x^4+2x^2+1} \\
    y'=\frac{x^2(x^2 +3)}{x^2(x^2+2+1)} \\
    y'=\frac{x^2+3}{x^2+2+1} \\
    y'=\frac{x^2+3}{x^2+3} \\
    y'=1 \\
    [/tex]
     
  14. May 22, 2013 #13

    Mark44

    Staff: Mentor

    Stop here. Expanding the denominator serves no purpose, and there is an error in your work.
    No.
    x4 + 2x2 + 1 ≠ x2(x2 + 2 + 1)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Basic Differentiation
  1. Basic Differentiation (Replies: 2)

Loading...