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Taking the derivative of e^3x

  1. Jan 11, 2009 #1
    If f(x) is the function given by f(x) = e3x + 1, at what value of x is the slope of the tangent line to f(x) equal to 2?

    I thought the derivative of e3x would be 3e3x because of the chain rule, but it doesn't appear to be correct. I know that ex is just ex, so is e3x just e3x?
     
  2. jcsd
  3. Jan 11, 2009 #2

    Dick

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    You were right the first time. You need the chain rule. (e^(3x))'=3*e^(3x).
     
  4. Jan 11, 2009 #3

    tiny-tim

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    Hi sonofjohn! :smile:
    It is correct … who says it isn't? :confused:
     
  5. Jan 11, 2009 #4
    Ahh thank you very much!
     
  6. Jan 11, 2009 #5
    The region bounded by y = e^x, y = 1, and x = 2 s rotated about the x-axis. The volume of the solid generated is given by the integral_____________.

    So this is volume problem and a disc/ washer method formula should work best. I was going to set the problem as:

    pi(antider)from 0-2 of (e^x-1)^2(1-1)^2 but I don't believe I will come out with 1-1 and 0 as the volume.
     
  7. Jan 11, 2009 #6

    Dick

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    You are rotating around the x-axis. At a given value of x what is the outer and inner radius of the washer?
     
  8. Jan 11, 2009 #7
    0 and 2, so they would be my bounds so would I rather set the problem up as, pi*antiderv*(e^x-1)^2
     
  9. Jan 11, 2009 #8

    tiny-tim

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    Hi sonofjohn! :smile:
    (have a pi: π :smile:)

    I don't actually understand all of your (e^x-1)^2(1-1)^2 …

    but you have a π(ex - 1)2, which is not the area of anything, is it? :wink:
     
  10. Jan 11, 2009 #9

    Dick

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    No, at say x=1 what is the inner and outer radius of the washer? What about at a general value of x? 0 and 2 are fine for the limits on the x integration. Now you just want to find the area of the washer. Correctly, this time. It's not pi*(e^x-1)^2.
     
  11. Jan 11, 2009 #10
    Ok so I should find the area of the washer. The area of the washer should be defined as e^x - 1 Also since I am using the dish washer formula, I should square both parts of the integration thus yielding:

    pi(anitderivative)(e^2x -1)
     
  12. Jan 11, 2009 #11

    Dick

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    Ok. Yes. The area is pi*(outer radius^2-inner radius^2).
     
  13. Jan 11, 2009 #12
    Thanks for the clarification!
     
  14. Jan 11, 2009 #13
    [​IMG]

    The 3rd problem on this page is difficult for me to understand. I don't understand what the x stands for when they are talking about the bounds in terms of integration. Do they mean the x-axis on the graph or possible x = 2 where at the x intercept?
     
  15. Jan 11, 2009 #14

    Dick

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    x is a point between 0 and 4. Any one. What you might notice is that G(x) is the area under the curve from 0 to x. H(x) is NEGATIVE of the area under the curve from x to 2 (because the integral is from 2 to x instead of from x to 2). Might this tell you something about G(x)-H(x)?
     
    Last edited: Jan 11, 2009
  16. Jan 11, 2009 #15
    So then the answer must be g(x) = h(x) - 2, because h(x) is positive and g(x) is negative. g(x) is also always going to be two less, because it goes from any point to 0, and h(x) only goes to 2.
     
  17. Jan 11, 2009 #16

    Dick

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    I can't really agree with you there, sonofjohn. Try this. What are G(1) and H(1)? Work them out from the picture.
     
  18. Jan 11, 2009 #17
    I see now that (d) cannot work. Subtracting 2 everytime from h will not yield an equal integral. Now I would like to say that g(x) = h(x+2) would work, but it doesn't seem plausible past h(1). Could G'(x) = H'(x+2) work? I don't even understand what it means.
     
  19. Jan 11, 2009 #18

    Dick

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    You didn't answer my last question. What are G(1) and H(1)? That should let you eliminate some possibilities.
     
  20. Jan 11, 2009 #19
    g(1) is -2 and h(1) is 1 so a,d,e cannot be the right answer .
     
    Last edited: Jan 11, 2009
  21. Jan 11, 2009 #20

    Dick

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    How did you get that? I get G(1)=2 and H(1)=(-1).
     
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