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Math Help - Making some progress on differentiation problem

  1. #1
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    Making some progress on differentiation problem

    f(x) = x^(3x)

    So far i used implicit differentiation and get
    (1/y)(y')=2x + 3ln(x)

    but I am way off,

    any help would be appreciated!
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  2. #2
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    Quote Originally Posted by frozenflames View Post
    f(x) = x^(3x)

    So far i used implicit differentiation and get
    (1/y)(y')=2x + 3ln(x)

    but I am way off,

    any help would be appreciated!
    Why are you way off?

     y = x^{3x}

     \ln(y) = \ln(x^{3x})

     \ln(y) = 3x.\ln(x)

     \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} 3x.\ln(x)

    You aren't far off it. Just use the product rule on the RHS.
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  3. #3
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    Quote Originally Posted by frozenflames View Post
    f(x) = x^(3x)

    So far i used implicit differentiation and get
    (1/y)(y')=2x + 3ln(x)

    but I am way off,

    any help would be appreciated!
    Express y as e^(3x*ln(x)) and use the chain rule.
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  4. #4
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    Quote Originally Posted by rn443 View Post
    Express y as e^(3x*ln(x)) and use the chain rule.

    i'm not sure how to solve for y'....because I would assume I would multiply both sides by "y" to get the result but the solution indicates that "y" is not supposed to be in the solution at all!
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  5. #5
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    Quote Originally Posted by frozenflames View Post
    i'm not sure how to solve for y'....because I would assume I would multiply both sides by "y" to get the result but the solution indicates that "y" is not supposed to be in the solution at all!
    Yes, but you know what y is from the original equation. Simply substitute it in.  y = x^{3x}

    So  \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{3x}} \frac{dy}{dx}

    So multiply through by  x^{3x} to get your equation  dy/dx = purely in terms of x.
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  6. #6
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    I submit the following and it shows up as incorrect..

    I may be missing something?

    x^(3x)[2x + 3ln(x)]
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  7. #7
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    Quote Originally Posted by frozenflames View Post
    I submit the following and it shows up as incorrect..

    I may be missing something?

    x^(3x)[2x + 3ln(x)]
    Yes. As I said before  2x + 3 \ln(x) is NOT the derivative of  3x. \ln {x}

    You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.

    y = x^{3x}

     \ln(y) = \ln(x^{3x})

     \ln(y) = 3x \ln (x)

     \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x))

    You must use the product rule on the RHS to differentiate.
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  8. #8
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    Quote Originally Posted by Mush View Post
    Yes. As I said before  2x + 3 \ln(x) is NOT the derivative of  3x. \ln {x}

    You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.

    y = x^{3x}

     \ln(y) = \ln(x^{3x})

     \ln(y) = 3x \ln (x)

     \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x))

    You must use the product rule on the RHS to differentiate.



    that gives me this: 3x(1/x)+3ln(x)

    still not correct
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  9. #9
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    Quote Originally Posted by frozenflames View Post
    that gives me this: 3x(1/x)+3ln(x)

    still not correct
    That's right.

     \frac{1}{y} \frac{dy}{dx} = 3 + 3\ln(x)

     \frac{dy}{dx} = (3+3\ln(x)) \times y

     \frac{dy}{dx} = 3x^{3x} (1+\ln(x))
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