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Math Help - implicit diff. with logs

  1. #1
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    implicit diff. with logs

    4x^3 + ln y^2 + 2y = 2x

    I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    4x^3 + ln y^2 + 2y = 2x

    I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
    It is implied that y is a function of x, Therefore, we treat y as composite (composed with x).

    \ln{y^2}=2\ln{y}

    \frac{d}{dx}[2\ln{y}]=2\frac{1}{y}\cdot\frac{dy}{dx}

    Do you see my point?
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    Quote Originally Posted by VonNemo19 View Post
    It is implied that y is a function of x, Therefore, we treat y as composite (composed with x).

    \ln{y^2}=2\ln{y}

    \frac{d}{dx}[2\ln{y}]=2\frac{1}{y}\cdot\frac{dy}{dx}

    Do you see my point?
    Just to verify that I got your explanation, would that mean the derivatives are;

    12x^2 + 2 y'/y + 2yy' = 2 ?

    I'm messing up somewhere in the calculations. I can't seem to get y(1 - 6x^2) / 1 + y which is the given answer.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    Just to verify that I got your explanation, would that mean the derivatives are;

    12x^2 + 2 y'/y + 2yy' = 2 ?

    I'm messing up somewhere in the calculations. I can't seem to get y(1 - 6x^2) / 1 + y which is the given answer.
    Everythin looks good except the third term on the left hand side. Should be...

    2y'

    not

    2yy'
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  5. #5
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    12x^2 + 2 y'/y + 2y' = 2
    2 y'/y + 2y' = 2 - 12x^2
    2y' (1/y + 1) = 2 - 12x^2
    y' (1/y + 1)= 1 - 6x^2
    y' = (1 - 6x^2) / (y + 1)

    Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.
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  6. #6
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    Quote Originally Posted by Archduke01 View Post
    12x^2 + 2 y'/y + 2y' = 2
    2 y'/y + 2y' = 2 - 12x^2
    2y' (1/y + 1) = 2 - 12x^2
    y' (1/y + 1)= 1 - 6x^2

    y' = (1 - 6x^2) / (1/y + 1)

    multiply numerator and denominator by y ...

    y' = y(1 - 6x^2)/(1 + y)

    Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.
    ...
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    12x^2 + 2 y'/y + 2y' = 2
    2 y'/y + 2y' = 2 - 12x^2
    2y' (1/y + 1) = 2 - 12x^2
    y' (1/y + 1)= 1 - 6x^2
    y' = (1 - 6x^2) / (y + 1)

    Where did I go wrong? The numerator's supposed to have a y in it, but my answer doesn't yield such a thing.
    \frac{d}{dx}[4x^3+\ln{y^2}+2y]=\frac{d}{dx}[2x]

    12x^2+\frac{2}{y}y'+2y'=2

    6x^2+\frac{1}{y}y'+y'=1

    \frac{1}{y}y'+y'=1-6x^2

    y'\left(\frac{1}{y}+1\right)=1-6x^2

    y'=\frac{1-6x^2}{\frac{1}{y}+1}

    y'=\frac{y-6x^2y}{1+y}
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  8. #8
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    Quote Originally Posted by Archduke01 View Post
    4x^3 + ln y^2 + 2y = 2x

    I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
    4x^3 + \ln{y^2} + 2y = 2x.

    \frac{d}{dx}(4x^3 + \ln{y^2} + 2y) = \frac{d}{dx}(2x)

    12x^2 + \frac{d}{dx}(\ln{y^2} + 2y) = 2

    12x^2 + \frac{d}{dy}(\ln{y^2} + 2y)\,\frac{dy}{dx} = 2

    12x^2 + \left(\frac{2y}{y^2} + 2\right)\,\frac{dy}{dx} = 2

    \left(\frac{2}{y} + 2\right)\,\frac{dy}{dx} = 2 - 12x^2

    \frac{dy}{dx} = \frac{2 - 12x^2}{\frac{2}{y} + 2}

    \frac{dy}{dx} = \frac{1 - 6x^2}{\frac{1}{y} + 1}.


    You could clean this up more if you wanted.
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    Thanks brosephs, I get it now. Damn, that 2 layered denominator was confusing.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    But if we're seeing y as a function, shouldn't we have to apply the chain rule?
    Sure! That's what we did!

    \frac{d}{dx}2y=(1)2y^{1-1}y'=2(1)y'=2y'
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  11. #11
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    Quote Originally Posted by VonNemo19 View Post
    Sure! That's what we did!

    \frac{d}{dx}2y=2y^{1-1}y'=2(1)y'=2y'
    XD yeah I realized that just after I posted. My badness.
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