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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- Dec 6th 2009, 02:46 PMArchduke01implicit 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? - Dec 6th 2009, 02:52 PMVonNemo19
- Dec 6th 2009, 02:59 PMArchduke01
- Dec 6th 2009, 03:10 PMVonNemo19
- Dec 6th 2009, 03:21 PMArchduke01
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. - Dec 6th 2009, 03:27 PMskeeter
- Dec 6th 2009, 03:27 PMVonNemo19
$\displaystyle \frac{d}{dx}[4x^3+\ln{y^2}+2y]=\frac{d}{dx}[2x]$

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

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

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

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

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

$\displaystyle y'=\frac{y-6x^2y}{1+y}$ - Dec 6th 2009, 03:27 PMProve It
$\displaystyle 4x^3 + \ln{y^2} + 2y = 2x$.

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

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

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

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

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

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

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

You could clean this up more if you wanted. - Dec 6th 2009, 03:40 PMArchduke01
Thanks brosephs, I get it now. Damn, that 2 layered denominator was confusing.

- Dec 6th 2009, 03:43 PMVonNemo19
- Dec 6th 2009, 03:44 PMArchduke01