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Checking a derivative

  1. Jun 13, 2004 #1
    Am I on the right track?

    Use the Summation Rule to find f ' (x) and simplify where possible:

    f ' (x) = [tex] \frac {\sqrt {x}} {3} - \frac {3} {\sqrt {x}} + \frac {2} {x^3} [/tex]

    = [tex] \frac {1}{3} (\frac {1}{2} x^{-\frac {1}{2}}) - 3(-\frac {1}{2} x^{-\frac{3}{2}}) + 2(-3{x^{-4}}) [/tex]

    = [tex] \frac {1} {6\sqrt {x}} + \frac {3} {2x^{\frac{3}{2}}} - \frac{6} {x^{4}} [/tex]
     
  2. jcsd
  3. Jun 13, 2004 #2

    arildno

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    Seems OK to me
     
  4. Jun 13, 2004 #3
    Derivative

    How about this one? Is this right?

    Using the summation rule, find f ’(x) and simplify where possible:

    f (x) = [tex] \frac {1}{(2x)^{3}} + \frac {\sqrt {x}}{2\sqrt [3] {x}} + \frac {x} {\sqrt {2}}[/tex]

    = [tex] \frac {1}{8}(-3)x^{-4}+ (\frac {1}{2})(\frac {1}{6})x^{-\frac {7}{6}}+\frac{1}{\sqrt{2}}[/tex]

    = [tex] \frac {-3}{8x^{4}} + \frac{1}{\sqrt{2}}[/tex] +1/12x^7/6

    (Sorry about the notation couldn't get the last part right)
     
  5. Jun 13, 2004 #4

    arildno

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    You should have
    [tex]\frac{1}{12}x^{-\frac{5}{6}}[/tex]
    rather than
    [tex]\frac{1}{12}x^{-\frac{7}{6}}[/tex]

    otherwise OK.
     
  6. Jun 13, 2004 #5
    Thanks..can you check this one?

    Using the summation rule, find f ’(x) and simplify where possible:

    F (x) = [tex] \frac {x^{4}+2x^{2}-3x}{4\sqrt{x}}[/tex]

    F ‘ (x) = [tex] (\frac{1}{4})(\frac{7}{2})x^{\frac {5}{2}}+ (\frac {1}{2})(\frac {3}{2})x^{\frac {1}{2}}-(\frac {3}{4})(\frac {1}{2})x^{-\frac{1}{2}}[/tex]

    = [tex] \frac {7}{8}x^{\frac{5}{2}}+\frac{3\sqrt{x}}{4}-\frac{3}{8\sqrt{x}}[/tex]
     
  7. Jun 13, 2004 #6

    arildno

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    It seems you are ready to move on to more challenging problems, ladyrae; you're mastering these concepts now :smile:
     
  8. Jun 13, 2004 #7
    Thanks...the next one is the Product Rule

    Use the product rule to find f ‘ (x) and simplify where possible

    f (x) = [tex] (x+e^{x})(3-\sqrt{x})[/tex]

    f ‘ (x) = [tex] (x+e^{x})(-\frac{1}{2}x^{-\frac{1}{2}})+(3-x^{\frac{1}{2}})(1+e^{t})[/tex]

    = [tex] -\frac{3}{2}x^{\frac{1}{2}}-\frac{e^{x}}{2x^{\frac{1}{2}}}+3+3e^{t}-x^{\frac{1}{2}}e^{t}[/tex]

    Is this right?
     
  9. Jun 13, 2004 #8

    Math Is Hard

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    Which book are you working out of? Does it not offer the solutions - or a solution manual that you can buy separately? If you are using Stewart Calculus, I can show you where to download some of the solutions manual chapters for free.(At least I think the link's still there).
     
  10. Jun 13, 2004 #9

    Math Is Hard

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    p.s. Wow! You are getting quite expert at the Latex formatting!!!
     
  11. Jun 13, 2004 #10
    Thanks

    I'm doing a calculus assignment...

    Can anyone check the last one for me?

    Thanks
     
  12. Jun 13, 2004 #11

    Math Is Hard

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    where did the e^t come from that I see in your derivative?
     
  13. Jun 13, 2004 #12
    terrible typo...i was reviewing another problem like it wrote down the wrong equation...

    f (x) = [tex] (x+e^{x})(3-\sqrt{x})[/tex]

    f ‘ (x) = [tex] (x+e^{x})(-\frac{1}{2}x^{-\frac{1}{2}})+(3-x^{\frac{1}{2}})(1+e^{x})[/tex]

    = [tex] -\frac{3}{2}x^{\frac{1}{2}}-\frac{e^{x}}{2x^{\frac{1}{2}}}+3+3e^{x}-x^{\frac{1}{2}}e^{x}[/tex]

    Is this right?[/QUOTE]
     
  14. Jun 13, 2004 #13
    2 left

    Use the quotient rule to find f ’(x) and simplify where possible

    Y = [tex] \frac{x^{3} + x} {x^{4}-2}[/tex]

    Y ‘ = [tex] \frac {(x^{4}-2) (3x^{2}+1)-(x^{3}+x)(4x^{3})}{x^{4}-2}[/tex]

    =[tex] \frac {-x^{6}-3x^{4}-6x^{2}-2}{x^{4}-2}[/tex]

    Is this right? How about the last one?

    Anyone...Thanks
     
  15. Jun 13, 2004 #14

    arildno

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    Post 12 is correct.

    The second line in 13 has the correct numerator, but you have forgotten to square the denominator.
     
  16. Jun 13, 2004 #15
    thanks

    Thanks..it should be

    [tex] \frac {-x^{6}-3x^{4}-6x^{2}-2}{(x^{4}-2)^{2}}[/tex]

    can this be simplified?
     
  17. Jun 13, 2004 #16
    last one

    Find the equation of the tangent line to the curve [tex] y = x^{3}-1 [/tex]
    at the point (-1,-2)

    y ' = [tex] \frac {d}{dx}(x^{3}-1) = 3x^{2} [/tex]

    [tex] m = 3x^{2} = 3 [/tex]

    [tex] (y-y_1) = m(x-x_1) [/tex]

    [tex] -3x+y-1 = 0 [/tex]

    or
    [tex] y = 3x + 1 [/tex]
     
  18. Jun 13, 2004 #17
    If you have a TI-89, it will find all of these derivatives for you. It's a very nice way to check your work.
     
  19. Jun 13, 2004 #18
    last one

    Find the equation on the tangent line to the curve y = [tex] x^{3}-1 [/tex]at the point (-1,-2)

    y ' = [tex] \frac {d}{dx}(x^{3}-1)=3x^{2}[/tex]

    [tex] m= 3x^{2} = 3 [/tex]

    [tex] (y-y_1) = m(x-x_1) [/tex]

    [tex] -3x + y - 1 = 0 [/tex]

    or
    [tex] y=3x+1 [/tex]

    is this right?
     
  20. Jun 13, 2004 #19
    sorry

    sorry for the repost..i didn't see page 2

    Anyone?
     
  21. Jun 13, 2004 #20

    Math Is Hard

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    You might also like to check out this site:

    http://www.calc101.com

    You can check derivatives here - although sometimes the steps it shows you will be more complicated than what you need to do to solve.
     
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