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Thread: Derivative of this logarithmic function.

  1. #1
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    Derivative of this logarithmic function.

    f(x) = ln5 + e25x - e3 +(7)/(cubed root of x) - 6

    f'(x) = (1)/(5) + e25x (25) - e3 - (7)/(3)(cubed root of x^4)

    is this correct?
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  2. #2
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    Re: Derivative of this logarithmic function.

    Hey XxXScorpionXxX.

    Since ln(5) is a constant the derivative of that term will be zero [as is another one].

    Try using the power rule noting that 1/cube_root(x) = x^(-1/3).
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  3. #3
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    Re: Derivative of this logarithmic function.

    (Beaten to it).
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    Re: Derivative of this logarithmic function.

    Quote Originally Posted by XxXScorpionXxX View Post
    f(x) = ln5 + e25x - e3 +(7)/(cubed root of x) - 6

    f'(x) = (1)/(5) + e25x (25) - e3 - (7)/(3)(cubed root of x^4)

    is this correct?
    no.

    $\ln(5)$ is just a constant so $\dfrac{d}{dx} \ln(5) = 0$

    same with $e^3$

    $f(x) = \ln(5) + e^{25x} - e^3 + \dfrac {7}{x^{1/3}} - 6$

    $f^\prime(x) = 25 e^{25x} -\dfrac 7 3 \dfrac{1}{x^{4/3}}$
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  5. #5
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    Re: Derivative of this logarithmic function.

    I see so 1/5 would be the derivative of ln5x then?

    And I did try to use the power rule for 7 divided by the cubed root of x. does that not comout to -7 divided by 3 times cubed root of x^4?
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  6. #6
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    Re: Derivative of this logarithmic function.

    thanks for the clarification romsek you're always a great help.
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  7. #7
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    Re: Derivative of this logarithmic function.

    Quote Originally Posted by XxXScorpionXxX View Post
    I see so 1/5 would be the derivative of ln5x then?
    NO!

    The derivative D_x(\log(5x))=\dfrac{5}{5x}=\dfrac{1}{x}.
    Last edited by Plato; Mar 17th 2017 at 02:01 PM.
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  8. #8
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    Re: Derivative of this logarithmic function.

    It's a bit wrong.

    It should be
    $$f'(x) = e^{25x} (25) -\dfrac{7}{3x\sqrt[3]{x}}$$
    $e^3$ and $\ln5$ should be removed as they are constants.

    and $f(x)=\dfrac{(7)}{\sqrt[3]{x}}\\f'(x)=\dfrac{-7}{3x^{4/3}}$
    Last edited by deesuwalka; Mar 21st 2017 at 11:02 PM.
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