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Math Help - why is this derivative wrong?

  1. #1
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    why is this derivative wrong?

    See attachment. It should be right but I'm wrong.
    Attached Thumbnails Attached Thumbnails why is this derivative wrong?-derivative.png  
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  2. #2
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    fine so far, but you haven't finished.

     \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)

     \displaystyle \frac{d}{dx} \left( \left[ 0.25 e^{-4t} \right]^{x^3}_0 \right)

     \displaystyle \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25e^0 \right)

     \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right)

    now differentiate with respect to x.

     0.25 \frac{d}{dx} \left( e^{-4x^3} \right)   - 0

    ...

    ...
    Last edited by SpringFan25; October 1st 2010 at 11:41 AM.
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  3. #3
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    also, according to wikipedia, the answer can be written down directly:
    Differentiation under the integral sign - Wikipedia, the free encyclopedia

    But you should use whatever method your teacher is expecting.


    Edit: Ignore this, not applicable here. (would be if the upper limit was x instead of x^3 )
    Last edited by SpringFan25; October 1st 2010 at 12:15 PM.
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  4. #4
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    Ok I understand what you did in the first four lines. I don't get the part where you say differentiate with respect to x and it goes "- 0 " at the end. Shouldn't it be -1 if the 0.25 was factored out like that?
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  5. #5
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    Yes on the factoring, but the derivative of 1 is 0.


     \frac{d}{dx} \left( 0.25e^{-4x^3} - 0.25 \right)

     0.25 \frac{d}{dx} \left( e^{-4x^3} - 1 \right)

     0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - \frac{d}{dx}\left(1 \right) \right)

     0.25 \left( \frac{d}{dx} \left( e^{-4x^3} \right) - 0 \right)


    Note There MAY be a mistake in my working somewhere else, as i dont think the answer that this will produce is consistent with what wikipedia says. However wikipedia math articles routinely have mistakes in them, so we will see ^^
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  6. #6
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    ok so the - 0 at the end isn't needed leaving us with just 0.25 d/dx * e^-4x^3 and isn't that what I wrote or should I ad "dt"?

    This would give (0.25 d/dx * e^-4x^3) * dt

    or is it because they ask for the derivative of d/dx that I must replace "t" by "x^3"?
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  7. #7
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    I dont see that in your screenshot. Even if it was what you meant, you haven't answered the question. you have to actually DO the differenciation.

    the answer is not d/dx "times" something. The answer is the derivative of 0.25 e^(-4x^3)


    NB
    Make sure you understand what the notation \frac{d}{dx}\left( f(x) \right) means. It does not mean that there is some fraction \frac{d}{dx} being multuiplied by a function, f(x).
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  8. #8
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    I don't understand.

    The way I read your line 4 is like this: 0.25 ( d/dx ( e^-4x^3 ) -0 )

    When I see this I read 0.25 and then the derivative of what's inside. If that is it then it should be e^-4x^3 * -12x^2 and we would get 0.25 * ( e^-4x^3 * -12x^2 ).
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  9. #9
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    and what does the textbook say the answer is?
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  10. #10
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    There is none. The tutor I went to see gave me this to work with:

    integral sign = S

    S (e^ax) * dx = 1/a * e^ax

    so I got: S (e^-4t)dt = -1/4 * e^-4t

    Then I think I sub t for x^3 but I'm not sure. I really don't know what to do after the sub. Do I apply F(b) - F(a) because I'm really lost. I'm so used to doing the F(b)-F(a) thing that when they ask this I just end up totally lost.
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  11. #11
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    I think that you are missing the point of the question.

    It is well known that if f is continuous on [a,b] and  \displaystyle g(x) = \int\limits_a^x {f(t)dt} then g'(x)=f(x).

    So with the aid of the chain rule  \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)=(3x^2)e^{-4x^3}
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  12. #12
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    0.25 * ( e^-4x^3 * -12x^2 ) was correct.

    You already did F(b) - F(a) right at the start to obtain this answer.

    Starting over:

    You want to find \displaystyle \frac{d}{dx} \left( \int^{x^3}_0 e^{-4t} dt \right)

    which i will write as \frac{d}{dx}something

    From the previous discussion, you should agree that
    "something" = 0.25e^{-4x^3} - 0.25

    you get this by doing the integral in the normal way, including the F(b) - F(a) step at the end.


    NOW start thinking about the derivative of "something", using the chain rule.

    differenciate \displaystyle \left(0.25e^{-4x^3}  - 0.25\right)

    =0.25 \times \frac{d}{dx}(-4x^3} \times e^{-4x^3}  -0 chain rule

    =0.25 \times -12x^2 \times e^{-4x^3}

    = -3x^2 \times e^{-4x^3}
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  13. #13
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    Okay just did this again from scratch, this is what I got:

    1/4 [(-e^-4x^3) + 1]

    What do you guys think?
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  14. #14
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    Quote Originally Posted by solidstatemath View Post
    Okay just did this again from scratch, this is what I got:

    1/4 [(-e^-4x^3) + 1]

    What do you guys think?
    Yes, that's the integration complete.
    Differentiate that and have another look at Plato's post.
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