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Math Help - Find the Derivative

  1. #1
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    Find the Derivative

    Ok, I have another problem that's bugging me pretty badly.

    Find the derivative:
    y=13t^2+40sqrt(t)+15/t

    I know that the derivative of 13t^2 = 26t
    but I'm unsure about what to do with the 40sqrt(t).
    Would it be something like:
    (.5)40x^-.5=1/20x?

    but then what about 15/t?
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  2. #2
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    Quote Originally Posted by bpdude View Post
    Ok, I have another problem that's bugging me pretty badly.

    Find the derivative:
    y=13t^2+40sqrt(t)+15/t

    I know that the derivative of 13t^2 = 26t
    but I'm unsure about what to do with the 40sqrt(t).
    Would it be something like:
    (.5)40x^-.5=1/20x?

    but then what about 15/t?
    Does it help if written in the form of exponents?
    y=13t^2+40\sqrt{t}+\frac{15}{t} = 13t^2 + 40t^{0.5}+15t^{-1}

    Spoiler:
    26t + \frac{20}{\sqrt{t}} - \frac{15}{t^2} = 26t^3+20t\,\sqrt{t}-15
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  3. #3
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    It does help a bit, but I'm still stuck
    I have:
    13t^2=26t
    40sqrt(t)=20t^-.5(?)
    15/t=15t^-1=-15
    so:
    26t-(1/20t)-15?

    Am I doing this right?
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  4. #4
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    Quote Originally Posted by bpdude View Post
    It does help a bit, but I'm still stuck
    I have:
    13t^2=26t
    40sqrt(t)=20t^-.5(?) e^(i*pi): Nay, because the square root is in the numerator it should be +0.5

    15/t=15t^-1=-15 e^(i*pi): No, the exponent is only on the t, the 15 is separate from t.
    so:
    26t-(1/20t)-15?

    Am I doing this right?
    ...
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  5. #5
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    e^(i*pi), could you possibly show me from the start how to go about solving this, so that maybe I can catch what I'm doing wrong? I'm not quite understanding your reply (I understand the sqrt(t) being +.5, but not the 15/t part).
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  6. #6
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    Quote Originally Posted by bpdude View Post
    e^(i*pi), could you possibly show me from the start how to go about solving this, so that maybe I can catch what I'm doing wrong? I'm not quite understanding your reply (I understand the sqrt(t) being +.5, but not the 15/t part).
    I did derive it in Post 2.

    Do you know how to derive

    <br />
y=13t^2 + 40t^{0.5}+15t^{-1}<br />

    This will be using the fact that \frac{d}{dx}(ax^n) = anx^{n-1} as you did for the first term
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  7. #7
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    Quote Originally Posted by bpdude View Post
    Ok, I have another problem that's bugging me pretty badly.

    Find the derivative:
    y=13t^2+40sqrt(t)+15/t

    I know that the derivative of 13t^2 = 26t
    but I'm unsure about what to do with the 40sqrt(t).
    Would it be something like:
    (.5)40x^-.5=1/20x?

    but then what about 15/t?
    y=13t^2+40t^{0.5}+15t^{-1}
    y'=13*2t^{2-1}+40*0.5t^{0.5-1}+15*(-1)t^{-1-1}
    y'=26t^{1}+20t^{-0.5}-15t^{-2}
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  8. #8
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    Wouldn't it be:
    26t+20t^-.5-15?
    Or maybe I'm missing something?

    I'm sorry if I am only being a 'headache' here.
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  9. #9
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    Quote Originally Posted by bpdude View Post
    Wouldn't it be:
    26t+20t^-.5-15?
    Or maybe I'm missing something?

    I'm sorry if I am only being a 'headache' here.
    Sry man, it wouldn't

    15/t=15*t^{-1} and (t^{-1})'=-1t^{-2}
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  10. #10
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    ialbrekht, YOU ROCK!

    So the overall solution was:
    y=26t+20t^-.5-15t^-2

    Thanks a lot guys! I really appreciate the responses!
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