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Help finding the second derivative

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x)= x^(3)e^(x)

    Find f'(x) in simplest form.
    Find f"(x) in simplest form.
    2. Relevant equations



    3. The attempt at a solution
    I found the first derivative to be: (using the product rule)
    f'(x) = (e^x)[3x^2] + (x^3)[e^x]
    f'(x) = 3x^(2)e^(x) + x^(3)e^(x)
    f'(x) = x^[2]e^[x](3 + x)

    I'm really not sure how to find f"(x) because of the "e". My initial thoughts would be to use the def. of derivatives on the problem, but then I get confused as to whether I should use the unfactored f'(x) or the factored version, which as far as I can tell would require the product rule. If anyone could help me get to f"(x) that would be great. Thank you!

    Also, the next part says to find the slope of the line tangent to the graph of f(x) at two different x values. Would I use f'(x) or f"(x) for this?
     
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 16, 2010 #2

    lanedance

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    to find f", just use the product rule twice ie,
    (a(bc))' = a'(bc) + a(bc)' = a'(bc) a(b'c + bc')

    for the 2nd part, which do you think represents the slope?
     
  4. Feb 16, 2010 #3

    rock.freak667

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    Write back f'(x) = x^[2]e^[x](3 + x) as ex(3x2+x3) and use the product law again.

    f'(x) or dy/dx gives the gradient function, which is the gradient of the tangent at a point x on the curve f(x).
     
  5. Feb 16, 2010 #4
    Okay, I did what you guys said and ended up with:

    f"(x)= x^[2]e^[x](15x^[2] + x + 3)

    I don't know if I did it right. What do you guys think?

    And now that I think about it it's obvious that f'(x) should be used for the slope.. Thanks guys!
     
  6. Feb 16, 2010 #5

    rock.freak667

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    I did not get that, post your working showing how you got that.
     
  7. Feb 16, 2010 #6
    f'(x)= e^[x](3x^[2] + x^[3])

    f"(x)= e^[x](3x^[2] + x^[3]) + e^[x]((6x)(x^[3]) + (3x^[2])(3x^[2]) <---I think I messed up in here

    = 3x^[2]e^[x] + x^[3]e^[x] + 15x^[4]e^[x]

    =x^[2]e^[x](15x^[2] + x + 3)

    I could've messed up anywhere and I'll never see it. Where did I go wrong?
     
  8. Feb 16, 2010 #7

    rock.freak667

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    u=ex

    v=3x2+x3

    now just find u(dv/dx)+v(du/dx)
     
  9. Feb 16, 2010 #8
    I ended up with 6xe^[x](x+2)

    In case I got it wrong, is the derivative of 3x^[2]+x^[3] 3x^[2]+6x? And how did you type the exponents in there?
     
  10. Feb 16, 2010 #9

    rock.freak667

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    use the X2 tag in the reply box.

    u=ex du/dx=ex

    v=3x2+x3 dv/dx=6x+3x2

    sub those back into the formula. You should get a different answer.
     
  11. Feb 16, 2010 #10
    How does xex(x2+6x+6) sound?
     
  12. Feb 16, 2010 #11

    rock.freak667

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    Much better.
     
  13. Feb 16, 2010 #12
    Thanks for the help! I caught my simple mistake there at the end.I was using the derivative of v twice in the equation instead of once... : /
     
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